The Schur test

Idea. The spectral radius of the symmetric matrix $A^TA$ equals $\|A\|^2$; bounding it by the *easily computable* norm $\|\cdot\|_1$ and then using submultiplicativity yields a bound in terms of row and column sums.

Step 1: $\|A\|^2=\rho(A^TA)$. By the problem on the spectral norm, $\|A\|^2=\lambda_{\max}(A^TA)$, and $A^TA$ is symmetric PSD so $\lambda_{\max}=\rho$.

Step 2: spectral radius versus an induced norm. $\|\cdot\|_1$ is the operator norm induced by the $\ell^1$-norm of $\mathbb{R}^n$, so it is submultiplicative, and the argument of the problem "Eigenvalues are bounded by the operator norm" (applied to a real eigenvector, since $A^TA$ is symmetric and has a real eigenvector $x$ for its largest eigenvalue $\mu$) gives $\mu|x|_1=|A^TAx|_1\le\|A^TA\|_1|x|_1$. So $\rho(A^TA)\le\|A^TA\|_1$.

Step 3: submultiplicativity.
$$\|A^TA\|_1\le\|A^T\|_1\|A\|_1=\|A\|_\infty\|A\|_1,$$
because the maximum column sum of $A^T$ is the maximum row sum of $A$.

Conclusion. $\|A\|^2\le\|A\|_1\|A\|_\infty$.

Example. For $A=\begin{pmatrix}1&-2\\3&4\end{pmatrix}$: $\|A\|^2\approx26.2$, while $\|A\|_1\|A\|_\infty=6\cdot7=42$. For a symmetric matrix $\|A\|_1=\|A\|_\infty$ and the inequality becomes $\|A\|\le\|A\|_1$.

Remark. This is the discrete form of the Schur test for boundedness of integral operators on $L^2$. It gives a cheap upper bound for $\|A\|$ with no eigenvalue computation. It can be quite far from sharp, e.g. for the all-ones matrix $J$ it gives $\|J\|^2\le n^2$ and in fact $\|J\|=n$ (sharp), but for a matrix with one large entry and many small ones it overestimates.