a simple and exact answer follows if $A$ is a unitary $n\times n$ matrix; then
$$P(u^H Av)=P(U_{11})={\rm constant}\,\times (1-|U_{11}|^2)^{n-2}$$
for $n\gg 1$ (and unitary $A$) the distribution of $\xi=|u^H Av|^2$ becomes exponential
$$P(\xi)=ne^{-n\xi}\;{\rm for}\;\; n\gg 1$$
for non unitary $A$, the answer will depend on its singular values $a_1,a_2,\ldots a_n$; I doubt that you can find a simple closed-form expression for any $n$, but for $n\gg 1$ I would think the same exponential distribution will appear, with $n$ replaced by $\sum_{i=1}^n a_i^2$.