I have tested by Matlab that for such a series of matrices $A_n \in \mathbb{R}^{n\times n}$ with $(A_n)_{i,j}=n-|i-j|$ (e.g.$A_3=\begin{bmatrix}3 &2 &1 \\ 2 &3 &2 \\ 1 &2 &3 \end{bmatrix}$), their smallest eigenvalues (denote as $(\lambda_n)_0$) are greater than 1/2, and $\lim\limits_{n\to\infty}(\lambda_n)_0=\frac{1}{2}$, but I don't know how to prove it. I only know that $$\det(A_n)=(n+1) *2^{n-2}$$ but I don't know it's useful or not.
In fact, I found this while solving the second problem of the 39th CMO held today :)
Note that $$ A_n^{-1}=\frac{1}{2(n+1)} \begin{pmatrix} n+2&-(n+1)&&&&1\\ -(n+1)&2(n+1)&-(n+1)\\ &-(n+1)&\ddots&\ddots\\ &&\ddots&\ddots&\ddots\\ &&&\ddots&2(n+1)&-(n+1)\\ 1&&&&-(n+1)&n+2 \end{pmatrix}. $$ By Gerschgorin disc theorem, all eigenvalues of $A_n^{-1}$ lie inside the interval $[0,2]$. It follows that $A_n$ is positive definite and $\lambda_\max(A_n^{-1})\le2$. Now let $v=\big(1,-1,1,-1,\ldots,(-1)^{n-1}\big)^T$. Then $v^TA_n^{-1}v=2(n-1)+\frac{1+(-1)^{n-1}}{n+1}$. It follows that $\lambda_\max(A_n^{-1})\ge\frac{v^TA_n^{-1}v}{v^Tv}=\frac{2(n-1)}{n}+\frac{1+(-1)^{n-1}}{n(n+1)}$. Hence $\lim_{n\to\infty}\lambda_\max(A_n^{-1})=2$ and $\lim_{n\to\infty}\lambda_\min(A_n)=\lim_{n\to\infty}\lambda_\max(A_n^{-1})^{-1}=\frac12$.
Remarks. This example of $A_n$ in the OP is fairly interesting. To stress the dependence of the vector $\big(1,-1,1,-1,\ldots,(-1)^{n-1}\big)^T$ on $n$, let us denote it as $v_n$ instead of $v$. In the answer above, we have shown that $\lim_{n\to\infty}\lambda_\max(A_n^{-1})=2$ and the Rayleigh quotient $\frac{v_n^TA_n^{-1}v_n}{v_n^Tv_n}$ approachs this limit when $n$ increases indefinitely. One may be tempted to think that $v_n$ is approximately an eigenvector of $A_n^{-1}$ corresponding to $\lambda_\max(A_n^{-1})$ when $n$ is large. One may also expect it to be approximately an eigenvector of $A_n$ corresponding to $\lambda_\min(A_n)$, and that $\lim_{n\to\infty}\frac{v_n^TA_nv_n}{v_n^Tv_n}$ exists and is equal to $\frac12$.
Surprisingly, none of these expectations is correct: the Rayleigh quotient $\frac{v_n^TA_nv_n}{v_n^Tv_n}$ is always equal to $1$! The vector $v_n$ is also far from being an eigenvector of $A_n$, in the sense that the elements of the entrywise division $(Av_n)\oslash v_n$ do not converge to $\frac12$. In fact, the first element of $(A_nv_n)\oslash v_n$ is equal to $\lceil \frac{n}{2}\rceil$, which diverges to infinity. The vector division $(A_n^{-1}v_n)\oslash v_n$ is more within expectation, though. It is equal to $(r_n,2,\ldots,2,r_n)^T$ where $r_n=\frac{n+2}{n+1}$ when $n$ is odd and $r_n=1$ when $n$ is even. All of its elements except the first and the last ones are equal to $\lim_{n\to\infty}\lambda_\max(A_n^{-1})=2$, but $\lim_{n\to\infty}r_n=1\ne2$.