There is a famous result by Schur, which tells us that if we define an operator $A$ on $\ell^2$ by setting $\langle A e_i,e_j\rangle=a_{ij}$ such that $\sup_j\lVert a_{ij}\rVert_{\ell^1_i}=\alpha$ and $\sup_i\lVert a_{ij}\rVert_{\ell^1_j}=\beta$ are both finite, then $A\in\mathscr{L}(\ell^2)$ is a bounded operator, with operator norm $\lVert A\rVert_{\mathscr{L}(\ell^2)}\le\sqrt{\alpha\beta}$.
Presumably, we cannot relax this condition by requiring an $\ell^2$-bound on the rows and columns and still expect $A$ to be a bounded operator. However, I have yet to find an example of how this may fail. Is there any such example that is easy to construct?
Furthermore, is a uniform $\ell^1$ bound the best we can do? Or can we generalize Schur's criterion to require and $\ell^p$ bound for some $1<p<2$?
Let $T$ be such that the $n$-th component of $Tx$ is $$(Tx)_n=\frac1n\sum_{j=1}^{n^2}x_j.$$ Say $x_j=1/N$ for $j\le N^2$, $0$ for $j>N^2$, so $||x||_2=1$. If $n\le N$ then $$(Tx)_n=\frac{1}{nN}n^2=\frac nN.$$And $$\sum_{n=1}^{N}\left(\frac nN\right)^2\ge cN\ne O(1),$$so $T$ is not bounded on $\ell_2$.
(Haven't looked at it carefully, but I bet that for $p>1$ just taking $(Tx)_n=\frac1n\sum_{j=1}^{n^p}x_j$ works...)
Edit: Yes, if $p>1$ and you define $T$ that way then, taking $x$ to be the $N$-th row of the matrix, if I did my sums correctly, you get $||Tx||_p^p\ge c N^{(p-1)^2}$.