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. 

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**EDIT**


[This journal article][1] 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 $A_x$).

  [1]: http://journals.cambridge.org/article_S1446181108000102