While there seems to be no explicit diagonalization, here is an idea that can get you started off toward a reasonable approximation.

To simplify notation, define $T=A_x$ be the original Toeplitz matrix (so I use $x$ 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.