Quick derivation
The Young-Laplace law states that
$$p-p_0 = \frac{2\gamma}{R}$$
whereas the equation of state of the ideal gas goes as
$$p = \frac{Nk_BT}{V}$$
Solving for $R$ and assuming that we are dealing with a spherical balloon ($V = \frac{4}{3}\pi R^3$, $A = 4\pi R^2$), and that the elasticity is described by a Hookean force (with equilibrium at zero size), $\gamma = \alpha A$,
$$\left(\frac{Nk_BT}{\frac{4}{3}\pi p}\right)^{1/3} = R = \frac{p-p_0}{8\pi\alpha}$$
To make the algebra simpler, I assume that $p_0 = 0$, so that we have $p\propto T^{1/4}$.
Slightly more rigorous derivation
For simplicity I'm going to assume that the pressure outside is zero. Adding non-zero pressure is trivial, though, but makes the equations a bit uglier.
Suppose we have a sphere filled with $N$ molecules of ideal gas, so that the partition function can be written as
$$\mathcal{Z} = \iint \mathrm{d}^{3N}p\ \mathrm{d}^{3N}r\ \ e^{-\beta(\mathcal{H}+\gamma A)}$$
So, we are left with
$$\mathcal{Z} = C V^N e^{-\beta\gamma A}$$
Now, minimizing the free energy with respect to $R$,
$$N\frac{A}{V} = \beta \partial_R(\gamma A)$$
Taking the rubber to be Hookean, $\gamma = \alpha A$, we finally have the size of the balloon:
$$R = \left(\frac{3N}{64\pi^2\alpha\beta}\right)^{1/4}$$
Now it is easy to compute the pressure,
$$p = -\left(\frac{\partial \mathcal{F}}{\partial V}\right)_A = \frac{N\frac{A}{V}}{\beta A} = \frac{N}{\beta V}$$
No surprise here; this is just the equation of state of the ideal gas. Plugging in the size ($V\leftarrow\frac{4}{3}\pi R^3$), we have $p \propto \beta^{-1/4} \propto T^{1/4}$.
I also wrote a simple Monte Carlo simulation (which could easily be extended to cover more general cases where the gas is non-ideal, say), and my numerical results agree with what I derived above.