The characteristic polynomial of $A_n^{-1}$ can be derived exactly to be \begin{align}\tag{1}\label{eq:1} P_n(\lambda) &= \frac{2^{1-n}}{n+1} + {}\\ &+\mathrm{Tr}\left[ \begin{pmatrix} \frac{1+n/2}{1+n} - \lambda & -\frac 1 4 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 - \lambda & -\frac 1 4 \\ 1 & 0 \end{pmatrix}^{n-2} \begin{pmatrix} \frac{1+n/2}{1+n} - \lambda & -\frac 1 {4(1+n)^2} \\ 1 & 0 \end{pmatrix} \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}}{n+1} \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}}{n+1} \, \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} 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} Insert this into the relevant factor of \eqref{eq:2b} and expand around $n=\infty$ to 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 $\Lambda_\mathrm{min}^{-1}=0.6755169\ldots$. Note that one can also get the large-$n$ corrections with this method.

The characteristic polynomial of OPs inverse $A_n^{-1}$ can be derived exactly to be (see, e.g., here) \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.