If $A$ is $n\times n$, then $$\frac1{\sqrt n}\|A\|_F\le\|A\|_1\le\sqrt n\,\|A\|_F,\qquad \|A\|_2\le\|A\|_F\le\sqrt n\,\|A\|_2.$$ More generally, if $A$ is $n\times m$, then $$\frac1{\sqrt m}\|A\|_F\le\|A\|_1\le\sqrt n\,\|A\|_F,\qquad \|A\|_2\le\|A\|_F\le\min(\sqrt n\,,\sqrt m)\,\|A\|_2.$$ To see that these inequalities are sharp, take respectively the matrices $$1_n\otimes \vec e^1,\quad \vec e^1\otimes 1_m,\qquad 1_n\otimes 1_m$$ for the first three inequalities. For the last one, complete the identity matrix by zeroes.
