Minimal covering of a $5 \times 5$ square with $T$ shaped tetrominoes.

This answer is based on Aqua's (now deleted) approach.

In the diagram there are three potential tetrominoes. There is no way to place a tetromino overlapping both red and green, or both blue and green. So if we want to have two tetrominoes overlapping all of these, one of the tetrominoes must overlap red and blue, and the only way to do that is to cover the marked $3\times 1$ rectangle.

But by rotating this picture, the same argument shows that if we have two tetrominoes and can't place another, we must also have a tetromino covering the horizontal $1\times 3$ rectangle which ends on the same square. These tetrominoes can't be the same (they cover too much space) but also can't be different (they overlap).

The LP relaxation of the integer linear programming formulation given here for this independent domination number problem yields objective value $8/3$, so the integer objective value is at least $\lceil 8/3 \rceil = 3$. By the way, the first several minimum values for an $n \times n$ square are: \begin{matrix} n &3 &4 & 5 &6 &7 &8 &9 &10 &11 &12 &13 &14 &15\\ \text{minimum} &1 &2 &3 &4 &6 &7 &9 &12 &14 &16 &20 &22 &25 \end{matrix}