My answer will be shamelessly Newtonian and Physics 101 in formulation. To start off the assumptions, I'm going to assume the air has no mass. To what extent is this valid? Air has about 1000x the density of other materials like rock and concrete, so we're looking at about that same volume ratio before the air mass becomes significant compared to the wall and as you'll see further into the calculations, this won't quite be the case until the object really is close to the size of the Earth.
The gravity at the surface of the balloon will be the following.
$$ g = \frac{G M }{R^2} $$
Here I have used the M variable to refer to the total mass of the wall. Now, this isn't the field that acts on the wall due to my prior arguments. Here's the part I was most unsure about: I divide this by two. Why? Well, the outside of the wall has $g$ act on it, but the inside of the wall has no gravitational field act on it at all (since I neglect the effect of the air). Average that out to get $1/2$. How does that translate into pressure? Introduce $\mu = \rho t$ where $t$ is the thickness of the wall, and you have the surface mass thickness in $kg/m^2$. This is what we need. Multiply that by the gravity and you have the same equation used on earth to find fluid head.
$$ P = \frac{1}{2} g \mu = \frac{G M}{2 R^2} \rho t$$
The point is that we already have a value of $P=14 psi$ that we wish to satisfy. For assumptions about density, $\rho$, my favorite approach is to assume it's made out of asteroid material with $\rho=1.3 g/cm^3$. Next, I'll introduce another easy equation, which is to multiply the mass thickness by the area to get mass.
$$M = 4 \pi \mu R^2 = 4 \pi t \rho R^2$$
These equations, with known density, by themselves can predict the shell thickness in what I call the "large limit". This assumed that the thickness is small relative to the total radius. So for any large space balloon made out of asteroid-density material the thickness is dictated by:
$$ t = \sqrt{ \frac{P}{2 G \pi \rho^2}} = 12.0 km$$
Since we know the thickness we may specify the radius or the mass. I thought it most appropriate to just say we have some given mass to work with. I took the mass of the asteroid 87 Sylvia, which is $1.5 \times 10^{19} kg$. Getting the rest is easy.
$$ R = \sqrt{ \frac{M}{4 \pi t \rho}} = 277.0 km$$
Yes, this is very big. However, the diameter is still about half that of Ceres. And 87 Sylvia is about the 18th largest by mass. Note that in the discussed configuration, the wall would occupy about 6.7% of the total volume.
Now I'm going to seaway into a different part of the answer where I ask "what if the balloon is fairly small?" We will start by defining $R$ to be the inner radius of the shell, which is the boundary of the air-filled region. To quickly get an answer, assume that $R\approx 0$, this forms the "small limit". You basically have a spherical asteroid and a negligible amount of air in the center. Integrate to find the fluid head, which will be set equal to 1 atmosphere.
$$ P = \int_0^t g(r) \rho dr = \int_0^t G \frac{4}{3} r \rho^2 dr = \frac{2}{3} G \pi \rho^2 t^2 $$
Now we get a definable limit for the smallest object we can make out of asteroid material that gets 1 atmosphere of pressure in its center.
$$ t = \sqrt{ \frac{3 P }{ 2 G \pi \rho^2 }} = 20.7 km $$
Obviously, this is larger than the previous large limit thickness, which is just due to geometrical factors. Now, how do we transition between the small limit and large limit values? We set up a more complex geometry, where the inner radius of the shell is $R$ and the outer radius of the shell is $R+t$. I struggled with this part of the problem a good deal, but I now have high confidence in this answer. To set it up, it's appropriate to say that the field within the rock is equal to the field you would have if the entire thing was solid (4/3 pi G rho r), minus the field you would get if the air were rock. This is using the superposition principle to subtract the rock in the middle which was "cut out".
$$ g(r) = g_{solid}(r) - g_{center}(r) = \frac{4}{3} \pi G \rho r + \frac{ G \rho \left( \frac{4}{3} \pi R^3 \right) }{ r^2 } \\
= \frac{4}{3} G \rho \pi t \frac{ \left( 3 R^2 + 3 R t + t^2 \right) }{ \left( R + t \right)^2 } $$
$$ P = \int_R^{R+t} g(r) \rho dr = \frac{2}{3} \pi \rho^2 G t^2 \frac{ 3 R+t }{ R+t} $$
This equation is easy to solve in terms of R, but not so easy to do in terms of t. It can also relatively easily be put in terms of P, M, and t.
I had graphs, but they were done when the equations were wrong, so I just wanted to get the math error corrected for now. Maybe I'll add more later.
