Let $T\in M_{n}(\mathbb{R})$ be a tridiagonal matrix. What can we say about operator norm $\|T\|_2$?
I'm asking this question because we know that if $T$ were only diagonal, then $\|T\|_2$ is the largest absolute value of any diagonal entry of $T$, as shown here, and so there might also a similar result for tridiagonal or $k$-diagonal matrices in general.
Edit. As user8675309 rightly points out, using Schur's test $\|A\|_2\le\sqrt{\|A\|_1\|A\|_\infty}$, we get $\|A\|_2\le3\max_{i,j}|a_{ij}|$. This bound is at least asymptotically tight: let $A_n$ be the $n\times n$ symmetric tridiagonal Toeplitz matrix with all entries on the three diagonals equal to $1$. Then $\lim_{n\to\infty}\|A_n\|_2=3$.
For a $(2k+1)$-diagonal matrix, Schur's test gives $\|A\|_2\le\min(n,\,2k+1)\max_{i,j}|a_{ij}|$.
(Old answer) Here is an obvious upper bound: since $A$ is tridiagonal, when we calculate $(AA^T)_{ij}$ as the sum $\sum_ka_{ik}a_{jk}$, at most three summands are involved. Also, as $AA^T$ is pentadiagonal, if we apply Gerschgorin disc theorem and the triangle inequality, we get $\rho(A^TA)\le(5)(3)\max_{i,j}|a_{ij}|^2$. Consequently, $$ \|A\|_2=\sqrt{\rho(AA^T)}\le\sqrt{15}\max_{i,j}|a_{ij}|. $$