Let $A\succeq 0$ and $M = A - \operatorname{diag}(A)$ be $M$ modified by setting the diagonal terms to zero. While $\operatorname{diag}(A)\succeq 0$, $M$ need not be positive or negative definite (consider $A$ as the all-ones matrix which has characteristic polynomial $(-1)^n\bigl(x - (n-1)\bigr)(x+1)^{n-1}$).
Can we bound $\|M\|$ in terms of $\|A\|$?
At least in the $A = \mathbf{1}\mathbf{1}^*$ example $\|M\|\leq \|A\|$ holds.
$\newcommand\diag{\operatorname{diag}}$This is true. Decompose $M = A - \diag A$ and pick $x$ of unit length so that $x^* M x = \|M\|$. Then $$ \|M\| = x^* M x = x^* A x - x^* \diag(A) x \leq \|A\| - x^*\diag(A) x \leq \|A\| $$ since $\diag(A)\succeq 0$. The more specific bound $$\|M\| \leq \|A\| - \min\{a_{ii}\}$$ holds similarly.