How many faces of a solid can one "see"?

You can always find a place from which to see at least half the faces.

To see why, start by considering a polyhedron with central symmetry. Imagine a viewpoint from which you don't see any lines as points or faces as lines (i.e. general position) and far enough away so that you can see all the faces whose normal points into your side of the half plane perpendicular to your line of sight. Then think about what you see from far enough away in the opposite direction. You can see all the faces from one side or the other and no face from both sides, so the symmetry says you see half each time.

Four of the five regular polyhedra have a center of symmetry. The tetrahedron does not: there's no place to put the origin that allows invariance under the map $x \to -x$.

Even without central symmetry, you see all the faces from one side or the other, so you see at least half from at least one side. Pyramids represent an extreme case. You can see all but one face from one direction and just one from the other, as @almagest points out in a comment.

Since the polyhedron has only finitely many faces, "far enough away" in the preceding proof does not have to be at infinity (though it may be pretty far). As @JohhHughes comments, if you put your camera close enough to any face that's the only face you'll see.

Note: the arguments work in all dimensions. They are particularly easy to visualize in the plane. (On the line they're trivial.)

As @almagest has pointed out, the absolute maximum number of faces you can see of a polyhedron with $n$ faces is $n-1$. This is achieved in the case of a right pyramid with a base and $n-1$ sides; if you view the pyramid from above the apex, you can see all the sides except the base. This is perhaps true for non-right pyramids and other shapes as well.

The absolute minimum number of faces you can see is 1, as you said: just place yourself (or the camera) arbitrarily close to any one face. As the polyhedron is convex, none of the faces will 'tower over' any one shape and you will see only one shape.

Both these bounds, however, are quite obvious and useless. As in the answer above, half the faces of a regular polyhedron with a center of symmetry can be seen from sufficiently far away. I would extend this to say that roughly half of the faces of a roughly regular polyhedron can be seen from sufficiently far away, where 'roughly' is an appropriate tolerance constant. I think it is illustrative to think of the sphere that completely circumscribes the polyhedron, of which you can obviously see exactly half. Maybe that half can be 'mapped' on to the polyhedron's faces.

Perhaps this paper which deals with the situation in a rather abstract way, and for higher dimensional polytopes might interest some: http://www-math.mit.edu/~rstan/transparencies/vis.pdf