$vAv$ does not make sense in general (unless $A$ is $1\times n$). What is common is, when $A$ is an $n\times n$ square matrix, the quantity $v^TAv$. In this case, if you consider the Frobenius norm for the vector and the induced operator norm for $A$, $$ |v^TAv|\leq \|A\|\,\|v\|^2. $$ If you want to use $\max|A_{ij}|$, then you can use the estimate $\|A\|\leq n\,\max|A_{ij}|$.