Isn't it simple?

$\overline{ABBA} =11(91A+10B) =11x$

As the number is a perfect square, $x$ is also a multiple of $11$

So, let $11x=121y$

As number is a perfect square, $y$ is a perfect square.

Let $121y=121m^2$

Putting values of $m$ as $3-9$ (as we have a 4 digit number) gives no number of the form $\overline{ABBA}$

So, no 4 figit perfect square is of form $\overline{ABBA}$

Therefore, required sum is $\boxed{0}$