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