tl;dr This is a really good question, and the implications for ground-based astronomers of various wavelengths are sobering.
Here's an example. The constellation described in SpaceX's 4,425 satellite constellation - what's the method to the madness? might even end up with more satellites now. From Phys.org's discussion of SpaceX's FCC applications in SpaceX details plans to launch thousands of internet satellites:
SpaceX has also indicated that it plans to deploy an additional 7.500 satellites that will operate at lower altitudes in order to boost broadband capacity in large population centers.
The simplest estimate of the number visible would be statistical. Let's say there are 5000 satellites that are uniformly distributed at an altitude of 1200 kilometers. This is not a correct model but it is a starting point for an order-of-magnitude estimate.
Seen from the center of the earth, each would occupy $4\pi/5000$ or about 0.0025 steradians, which rhymes with "square radians". So that's about 8.25 square degrees (as seen from the Earth).
Standing on the Earth and looking up at a shell of orbits at an altitude of 1200 km, and then only looking at the section within 45 degrees of your zenith (at half-way up or more from the horizon), you are looking at the intersection of a cone drawn from the center of the Earth with a half-angle of about 8.6 degrees, or about 232 square degrees.
That means that in that 45 degree half-angle cone you are looking in, there would be, on average, twenty seven (27) SpaceX satellites looking back at you!
If it's not close to midnight, a good share of them will be sunlit.
However, they are much farther than most satellites that we actually notice, and that means they may be dimmer and slower than the ones we most often notice and remember. Depending on how they are designed, they might not reflect all that much sunlight most of the time. Those details are very design dependent, and SpaceX is likely to be thinking about this already.
Telescope users:
Every one-degree FOV view would have a 10% chance of having one of these in it, on average at any moment, moving at about 0.35 degrees per second at the zenith and about 0.25 degrees per second at 45 degrees from the zenith. slowing down even more closer to the horizon.
So unless I'm doing something wrong here, a thirty second exposure through a telescope using low magnification with a 1 degree FOV would probably have a SpaceX satellite in it if you're within +/- 50 degrees latitude!
This is an average, your milage will vary depending on latitude.
Observatory users:
This would be a good question for Astronomy Stackexchange. I can say that for some exposures meant to record both dim and bright objects in the same field or on the same CCD array, CCD frames may be read out many times during the exposure. This might allow for subtraction of satellite streaks. Of course, there are plenty of satellites now, so this must already be done regularly. I think that's quite an interesting question, and it should be asked there if it hasn't been already.
Now, instead of random, let's look at the orbits for this example. Study the drawings and data below, then please go read this excellent answer explaining how this proposed configuration works conceptually!
The "concentration" in our 45 degree cones will be lower when we are near the equator, perhaps 15 or 20, but really spike at higher latitudes, with perhaps fifty or even more when we are between 45 degrees and 60 degrees latitude!
above x2: screen captures from SpaceX non-Geostationary Satellite System Attachment A - Technical Information to Supplement Schedule S.
