On Earth, if you go high enough you enter orbit and are effectively in free-fall.
That's not quite right. If you go
fast enough, if your velocity parallel to the Earth's surface is great enough that the curvature of your fall toward the Earth exceeds the curvature of the Earth's surface, then you are in orbit, perpetually falling toward the planet but never hitting it. (Some, myself included, like to paraphrase Douglas Adams and say that the secret to orbit is knowing how to throw yourself at the ground and miss.) Conversely, if your lateral movement isn't fast enough, you're still in free fall, being pulled toward the Earth by its gravity, but your falling trajectory will eventually intersect with the surface. There's no cutoff point where gravity suddenly stops pulling on you. It decreases as the inverse square of your distance.
So altitude really has nothing to do with it, except that the speed you need to maintain orbit gets smaller the farther out you go. Theoretically, if you were on a planet with no atmosphere and a completely flat surface, you could maintain an orbit one meter above its surface if you were going fast enough.
Given that a Dyson Sphere doesn't have the curvature of the Earth to keep objects in 'near sphere orbit', how far up would an object have to be before it went into orbit around the host star as opposed to plummeting back down to the sphere floor?
Well, first off, let's remember that at any point within a hollow sphere, the gravitational force exerted by the sphere's mass cancels out to zero. This is one of the fundamental flaws with the fictional conceit of a Dyson shell inhabited on the inner surface. Realistically, anything placed on the inner surface of a Dyson shell would be subject only to the star's gravitational pull.
Now, as various posters have discussed above, the usual fix for this is to assume some kind of artificial gravity exerted by the material of the shell. I'm not sure that would work. If the AG mechanism behaved like any standard force, i.e. dropping off as the inverse square of the distance, then if there were AG plating over the entire surface, its effects would cancel out everywhere within the sphere just as real gravity would, and it simply wouldn't have any effect. Even if it were some kind of magnetic pull, I think the same kind of shell-theory equations apply to EM fields within a charged hollow sphere.
So that leaves us with the kind of inexplicably short-ranged artificial gravity we see in
Star Trek, where you can be standing on top of a starship's saucer and be in free fall even though there's gravity plating just three or four meters below you. I've postulated (in my novel
Star Trek Titan: Orion's Hounds) that this could be the result of using virtual gravitons that decay quickly, so that their effects aren't felt beyond a given distance. It seems that this might be the only way to avoid the cancellation effect that produces uniform free-fall conditions within the shell.
So in either case, if you're any distance above the surface at all, you're going to be unaffected by the shell's gravity and will only be subject to the star's gravity. So the question isn't how high you need to get. It's how fast you're going. Remember, there's no cutoff distance for normal gravity. If you're anywhere inside the shell, and outside the effect of this conjectural short-range artificial gravity, then the star's real gravity will be pulling you in toward it. And that means you need to move fast enough laterally to avoid falling into the star -- yet not so fast that your orbital curve has a greater radius than the curve of the Dyson shell, or you'll crash into it.
