The characteristic polynomial of OPs inverse $A_n^{-1}$ can be derived exactly to be (see, e.g., [here][1])
\begin{align}
p_n(\lambda) &= \det(A_n^{-1}-\lambda I_n)=\frac{2^{1-n}}{1+n} + {}\\
\tag{1}\label{eq:1}
&+\operatorname{Tr}\left[ 
\left(\begin{smallmatrix}
\frac{1+n/2}{1+n} - \lambda & -\frac 1 4 \\ 1 & 0
\end{smallmatrix}\right)
\left(\begin{smallmatrix}
1 - \lambda & -\frac 1 4 \\ 1 & 0
\end{smallmatrix}\right)^{n-2}
\left(\begin{smallmatrix}
\frac{1+n/2}{1+n} - \lambda & -\frac 1 {4(1+n)^2} \\ 1 & 0
\end{smallmatrix}\right) \right].
\end{align}
Using the substitution $\cos\varphi = 1-\lambda$ this simplifies to
\begin{align}\tag{2a}\label{eq:2a}
p_n(\varphi) 
&= \frac{2^{1-n}}{1+n} \left[
1 + \cos(n \varphi) - n \sin(n \varphi) \tan\left(\tfrac\varphi 2\right)
\right] \\ \tag{2b}\label{eq:2b}
&=\frac{2^{2-n}}{1+n} \, \cos^2\left(\tfrac{n\varphi} 2\right)
\left[
1 - n \tan\left(\tfrac{n\varphi} 2\right)\tan\left(\tfrac{\varphi} 2\right)
\right] .
\end{align}
We observe that half of the eigenvalues of $A_n^{-1}$ (and $A_n$) are trivial ($\varphi_k = 2\pi k/n$ with odd $k$ fulfilling $0<k<n$), while the other half are roots of the last term in \eqref{eq:2b}. The minimal eigenvalue $\lambda_\mathrm{min}(n)$ of $A_n^{-1}$ belongs to the latter set.

To get the asymptotic smallest eigenvalue $\lambda_\mathrm{min}(n) \sim \Lambda_\mathrm{min} n^{-2}$, we expand
\begin{align}\tag{3}\label{eq:3}
\varphi=\arccos(1-\lambda) = \sqrt{2\lambda}+O(\lambda^{3/2}).
\end{align}
Inserting this into the relevant factor of \eqref{eq:2b} and expanding around $n=\infty$, we get
\begin{align}\tag{4}\label{eq:4}
p_\infty(\Lambda) = 
1 - \sqrt{\frac \Lambda 2} \tan\left(\sqrt{\frac \Lambda 2}\right).
\end{align}
The numerical solution of $P_\infty(\Lambda_\mathrm{min})=0$ is 
$\Lambda_\mathrm{min}=1.4803477\ldots$, and we get the result
$\Lambda_\mathrm{min}^{-1}=0.6755169\ldots\,.$

Note that one can also get the large-$n$ corrections with this method.


  [1]: https://arxiv.org/abs/0712.0681