TL, DR
Using a lattice structure and steel we arrive at ~ 2800 km across, a few hundred km smaller than earths moon. only about 2% of the volume are actually filled.
The problem with thin shells
You can't make a shell with an arbitrarily thin shell, as thin shells will buckle, long before the stress reaches the compressive strength. Approximately half of the other answers don't take this into account and are wrong.
I interpret the spirit of the question: Can there be a porous, honeycomb structure with lots of internal volume?
Luckily, lattices (3D-honeycombs) have been investigated and can be approximated as a bulk material. In a built structure - unlike a planet, which is by definition in hydrostatic equilibrium - the upper layers can carry their own weight and need not exert pressure on the lower layers. This is why thin shells are so popular in the answers here.
Approach 1
We can approximate a sphere as series of layers, each designed to carry their own weight. Analytically, we treat them as infinitely thin, however as we assume lower layers we need not worry about buckling. It might make sense structurally to transfer some load downward, but then the math gets more hairy. The upside of this approach is that the interior of the structure is less filled with stuff.
What we need to do is take the formula for the stress in a thin shell under it's own weight, and modify for the lower density and strength of the lattice.
The Force $F$ acting on piece of size $A$ of the outermost shell with thickness $t$ depens on the mass of the whole structure and is given with:
$$F = A \rho t G \rho \frac{4}{3} \pi$$
Note how $\rho$ goes into the left part - weight of the shell-element - and right part - total gravity - of the right hand side. If we move $A$ to the left by division we arrive at a sort of pressure acting on the shell. The hoop stress in pressure vessel is given by $\sigma = \frac{Pr}{2t}$, this relationship holds here too, its just compressive stress (not tensile stess). For the stress in our outtermost shell we arrive at:
$$\sigma = \frac{2}{3}G \rho^2 r^2 \pi$$
The paper mentioned above gives the following link between density and yield strength:
We see that by picking a pyramidical lattice and 0.02 denstiy - meaning 2% of the available volume is filled by the material - we get about 1% of the yield strength. 439
Presumably the pyramidical lattice looks something like this:
Now it's just entering numbers for your favorite material, with my favorite material (concrete), these are a compressive strength of 20-80 Mpa and a denstiy of about 2600 kg/m³. We'll assume 20 Mpa to account for a safety factor and arrive at 727 km radius, and 84 million tons. This is almost twice as large as Ceres, but far lighter.
Now, how about mild steel? Values for compressive strength for steel are hard to find, as metal rods under compression usually fail in shear or buckling. However the strength is higher than the tensile strength. So we assume a high strength alloy with a yield strength of 690 MPa and a density of 7.8 g/cm³. For funs sake, no safety factor is assumed. With these values I arrive 1426 km radius and 1.8 billion tons. As above, the surface gravity is on order of magnitude of 10^-5 m/s² - not enough to hold an atmosphere. The radius is only 300 km less than earths moon!
Why are these so small? Remember, the outtermost layer has to carry it's own weight. This means that at any circular hoop the weight of one hemisphere presses against the other, causing compressive stress. The weight scales (assuming constant gravity) with the square of the radius, the are only linearily. The same reason pressure vessels and pipe become weaker against internal pressure with larger size and constant wall thickness.
Note that my approach rests on assuming thin shells and in practice the thinnest thinkable shell in a lattice structure is at least as strong as a truss is long, this may introduce major errors - I simply don't know & don't know how to solve without doing a finite element analysis (which I also don't know how to do).
A view from the inside
From the image above we see that one cell of our lattice has 24 outer trusses and 12 inner trusses, but half of the outer trusses "belong" to other cubes so for the following we assume a total of 24 trusses. 441