How would I find the molarity/concentration of a gas using pV = nRT?
Supposing the dissociation process is isochoric, the total volume $V$ required for calculation of molarity at equilibrium of $i$th component $c_{\mathrm{eq},i}$

$$c_{\mathrm{eq},i} = \frac{n_i}{V}$$

can indeed be found from the ideal gas law using initial amount of undissociated sulfur trioxide:

$$V = \frac{n_0(\ce{SO3})RT}{p} = \frac{\pu{1.00 mol}\cdot\pu{8.21e-2 L atm K-1 mol-1}\cdot\pu{1000 K}}{\pu{1 atm}} \approx \pu{82.1 L}$$

You have found amounts of substances at equilibrium correctly, the rest is just math:

$$c_\mathrm{eq}(\ce{SO3}) = \frac{\pu{0.25 mol}}{\pu{82.1 L}} \approx \pu{3.0e-3 mol L-1}$$

$$c_\mathrm{eq}(\ce{SO2}) = \frac{\pu{0.75 mol}}{\pu{82.1 L}} \approx \pu{9.1e-3 mol L-1}$$

$$c_\mathrm{eq}(\ce{O2}) = \frac{\pu{0.375 mol}}{\pu{82.1 L}} \approx \pu{4.6e-3 mol L-1}$$