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 $T \ll \alpha^{-1}$ 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.
