Upper-bound for nuclear norm of $A \circ (v \otimes v)$ in terms of operator norm (or nuclear norm) of matrix $A$ and $L_\infty$-norm of vector $v$.
As initially observed by user Ben Grossmann in the comments, one has $\|B\|_\star \le r_2 r_3^2$.
Indeed, if $D = diag(v)$, then $B = DAD$ and so $$ \|B\|_\star = tr(B) = tr(DAD) = tr(AD^2) \le tr(A)\|D\|_{op}^2 = \|A\|_\star\|D\|_{op}^2 \le r_2 r_3^2, $$ where we have used the fact that $A$ and $D$ are symmetric psd matrices (thus so is $B$).
Moreover, this inequality is tight as can be seen by taking $v=1_n:=(1,1,\ldots,1) \in \mathbb R^n$, so that $B=A$.