Interesting question. Well it will be far from Earth well beyond the Moon which will make it a much fainter star than if it is in LEO. But it won’t move across the sky like the stars, so if you saw it with a conventional telescope it would move differently. In summary I make it’s magnitude between magnitude 17 and 18, with a “back of the envelope calculation”, and its apparent motion would be describing a circle around the antisolar point. This would make it easily visible within moderate sized (though fairly large) amateur telescopes. I make it that a 22 inch telescope should see it. But this is all rough first ballpart estimates. If I made no major mistake, I’d expect it to be accurate within a magnitude or two. Do take a look, and see if you think I got it right and do say if you know a better way to estimate it.

Here is its orbit.

And another take on it with more details

Both from: Orbit - JWST/NASA

So it slowly orbits around the L2 position, once every six months, at a distance from L2 similar to the distance of Earth from the Moon. It orbits there in order to keep all the bright objects in the sky for it, Earth the Moon and the Sun in the same direction so its heat shield can protect it from the light from them all. It’s designed to be so cold that even the light from the Moon, which would only be a thin crescent moon as seen from there, still needs to be shielded as it would have some warming effect.

It’s distance from Earth is 1.5 million kilometers. That’s 30,000 times the distance to a satellite in LEO passing overhead. But the brightness goes down as the square of the distance. So it would be 30,000^2, or 900,000,000 times fainter. That makes it around 22.5 magnitudes fainter. (100 times fainter equates to 5 magnitudes). By way of example, the ISS at its brightest is magnitude -5.6 ISS - Satellite Information

So the ISS would be about magnitude 16.9. Its solar panels have an area of 2,500 square meters as seen from Earth. ISS Facts!

The James Webb heat shield would be quite a large object 21.197 m x 14.162 m

That makes its surface area is 21.197 m x 14.162 / 2 (using the formula for the area of a kite). Or about 150 square meters. So it would be about 16.666 times fainter or a bit over 3 magnitudes fainter - that’s not taking account of the difference that the solar panels absorb light and convert them into power so optically dark while the heat shield re-radiates as much of the light as possible. So it would be much brighter. So that makes it brighter than 20 magnitudes, and isn’t really a very good upper bound on the apparent magnitude, but it’s a start.

Credit: STScI The James Webb Space Telescope

It’s designed as a radiator, not a reflector, so the heat I think would be radiated in all directions on the side towards the sun (though of course almost not at all towards the telescope). Temperature 85 C. It’s going to be in sunlight all the time as its distance from the L2 point is about the same distance as the distance of the Moon from Earth. Almost never obscured by the Moon. I suppose it could be, but only if the Moon happens to pass in front of it at the same time it is in the right part of its orbit to line up behind the Moon. So if it does, it would surely be a rare event.

So it would be a pretty much, or completely, constant feature of the night sky. I’d be surprised if it was bright enough to be naked eye visible, as much larger asteroids are not naked eye visible at that distance. Naked eye visibility is down to magnitude 6 or rarely 7 for keen eyed experienced observers from remote dark sites.

It would move in the sky in a way that would make it clear it couldn’t be anything else because it would slowly orbit the antisolar point in the sky as seen from Earth. But very slowly. So you’d need to watch it for a while to be sure it is moving and then to confirm that it is in an apparent orbit around the antisolar point as seen from Earth. But I can imagine amateur astronomers doing that and superimposing photographs to show it in many different positions tracing out a circle in the night sky with the other stars as short streaks, or a cycloid type pattern if done relative to the other stars shown stationary..

The average sun side temperature is not that different from the Moon. You’d expect it to be a bit warmer indeed, so we can use the Moon as a way to calculate a lower bound for its brightness. The global surface temperatures of the Moon as measured by the Diviner Lunar Radiometer Experiment

So the Moon has magnitude -12.6 What is apparent magnitude?

It’s cross sectional surface area is PI*(3,474/2)^2 = 9500000 square kilometers very approximately.

Compared to 150 square meters, or around 0.00015 square kilometers for the James Webb heat shield.

So the ratio of brightnesses is 9500000 /0.00015 or about 63,333,333,333. Or about 27 magnitudes. Giving it a magnitude of around 14.4 at the distance of the Moon.

But it’s (1.5 million / 384,400) times further away. Or about 3.9 times further away. So in brightness, around 3.9^2, or around 16 times fainter, very approximately. Or about 3 magnitudes fainter.

That would make it magnitude about 17.4.

This is a very rough calculation, assuming an albedo similar to the Moon. But surely it has higher albedo than that, probably far higher, as the Moon has albedo similar to dark asphalt. It might be similar in brightness even to the larger ISS with that 16.9 magnitude. But I think we have a reasonable ballpark figure to within a magnitude or two.

BTW I know I made many approximations in the calculations above, sometimes rounding up, sometimes rounding down. You could go through and use the exact values and only round at the end. There’s a theorem that if you do that, you get a result that is usually surprisingly close to the value you get if you round at each stage - the errors tend to compensate so long as you round randomly, e.g. round to the nearest value at some level of precision. So, someone could try a more precise calculation for sure. Or there may be a more direct way to work out its magnitude. But meanwhile, for this rough calculation I’m not sure if it is worth doing that.

So, basically yes, it would be a feature of the sky in our more sensitive telescopes and sky searches, e.g. searches for asteroids etc. For instance it would be well within the limiting magnitude of the Pan Starrs asteroid survey with limiting magnitude 24. Pan-STARRS - Wikipedia

But it would also be visible in quite small telescopes. Even the ISS at that distance would be visible in a 45 cm (18 inch) telescope Limiting Magnitude. The James Webb, if we got this calculation right, would be visible in a 22 inch (56 cm) telescope. Or a bit under 2 feet in diameter. So it would be a reasonably easy to spot object I think even in moderate sized amateur telescopes. The size that’s small enough that keen astronomers are able to transport them in cars and assemble at star parties.

So, if this is right, I expect it may be a thing that amateur astronomers might do, to photograph James Webb and track its position through the sky, and the loops it makes around the apparent antisolar point (which itself slowly moves across the background of the distant stars). They could superimpose those images either in such a way as to keep the stars fixed, or to keep the antisolar point fixed and let the stars drift slightly.

Any corrections, or if you see an easier way to work this out, do say. It’s just a first back of the envelope sketch. But unless I’ve made a major mistake, I expect it is right within a magnitude or two.