Yes: for white noise perturbations of the 1D dynamical system $\dot{x}=b(x)$, the action functional is $S(\phi)=\frac{1}{2} \int_0^T |b(\phi(s))-\dot{\phi}(s)|^2 ds$ for $\phi \in H^1$ and otherwise infinity. The LDP says that $\sigma^2 \log(P(X_\cdot \in A))$ is asymptotically between $-\inf_{f \in \mathrm{Int}(A)} S(f)$ and $-\inf_{f \in \mathrm{Cl}(A)} S(f)$ as $\sigma \to 0$.

In your case $A$ is the exterior of a ball in $C([0,T])$ centered at zero, so these two are the same, and so the problem is (presumably) to compute $I=\inf_{\phi \in C([0,T]) : \| \phi \|_\infty \geq z,\phi(0)=0} \frac{1}{2} \int_0^T |\dot{\phi}(s)+\alpha \phi(s)|^2 ds$. Then the logarithmic asymptotic for your quantity is $e^{-\sigma^{-2} I}$. I suspect the minimizer is attained by putting $\phi(T)=z$ (giving yourself the maximum possible amount of time to get to a magnitude of $z$) so that by the Euler-Lagrange equation $\phi(t)=z \frac{\sinh(\alpha t)}{\sinh(\alpha T)}$ is a minimizer; you should check this guess though. If I'm right then that means the minimum of $I$ is $\int_0^T \left ( z \alpha \left ( \frac{\cosh(\alpha s) + \sinh(\alpha s)}{\sinh(\alpha T)} \right ) \right )^2 ds = \alpha z^2(\coth(\alpha T)+1)$. This is large compared to $z^2$ for small $T$ but quickly settles toward $2\alpha z^2$; unsurprisingly this means that in the long run you should expect to see maximum deviations from zero on the order of $\frac{\sigma}{\sqrt{\alpha}}$.

The situation is not much different in higher dimensions provided the noise covariance matrix is constant and nonsingular (even if it isn't isotropic).

Cf. *Random Perturbations of Dynamical Systems* by Freidlin and Wentzell, chapter 3 in the 3rd edition for more details.