While there seems to be no explicit diagonalization, here is a simple idea to obtain an approximation.
To simplify notation, define $T=A_x$ to be the original Toeplitz matrix (I'm using $x$ instead of $\alpha$ to save typing).
After some playing around, it can be seen that the inverse enjoys remarkable structure, namely, $$ T^{-1} = \frac{1}{1-x^2} \begin{bmatrix} 1 & -x & \cdots & \cdots & 0\\\\ -x & 1+x^2 & -x & \cdots & 0\\\\ & \ddots & \ddots & \ddots\\\\ 0 & \cdots & &1+x^2 & -x\\\\ 0 & \cdots & &-x & 1 \end{bmatrix} $$
Consider, therefore, the following Toeplitz matrix \begin{equation*} M := \begin{bmatrix} 1+x^2 & -x & \cdots & \cdots & 0\\\\ -x & 1+x^2 & -x & \cdots & 0\\\\ & \ddots & \ddots & \ddots\\\\ 0 & \cdots & &1+x^2 & -x\\\\ 0 & \cdots & &-x & 1+x^2 \end{bmatrix}, \end{equation*} for which one has closed form eigenvalues and eigenvectors, given by
\begin{equation*} \lambda_k = (1+x^2)-2x\cos\left(\frac{k\pi}{n+1}\right),\quad 1 \le k \le n, \end{equation*} and \begin{equation*} v_{ik} = \sin\left(\frac{ik\pi}{n+1}\right),\quad 1 \le i \le n, 1 \le k \le n. \end{equation*} These eigenvalues (after scaling by $1-x^2$) and eigenvectors may be approximately substituted for those of $T^{-1}$.
Simple experimentation reveals that the eigenvalues of the matrix \begin{equation*} M' := \begin{bmatrix} 1 & -x & \cdots & \cdots & 0\\\\ -x & 1+x^2 & -x & \cdots & 0\\\\ & \ddots & \ddots & \ddots\\\\ 0 & \cdots & &1+x^2 & -x\\\\ 0 & \cdots & &-x & 1 \end{bmatrix} \end{equation*} satisfy $$|\lambda(M) - \lambda(M')| \le 4x/n,$$ where the bound can be made tighter by closer analysis (note that $\lambda(M) \ge \lambda(M')$ also holds). Similar results can also be shown for the eigenvectors, but I haven't had the time to prove that.
EDIT
This journal article shows how to compute eigenvalues and eigenvectors for matrices that look like $T^{-1}$ above. In particular, it implies for example, that the explicit eigenvalues of $T^{-1}$ above are given by
$$\lambda_k = 1 - 2x\cos\theta_k,$$
where, for $k=1,\ldots,n$, the angle $\theta_k$ is a root of $$\sin(n+1)\theta + x^2\sin(n-1)\theta - x\sin(n\theta).$$
Formulae for eigenvectors can also be found in terms of the $x$ and $\theta_k$ as stated above.
However, it seems that in our case, we'll have to numerically solve for $\theta_k$. Modulo that, I guess, this is as close as we'll get to explicit eigenvalues and eigenvectors of $T^{-1}$ (and thereby of $A_x$).