Answer: $n = 3k$, $n = 3k + 1$.
It is clear that any rectangle with one of the sides divided by $3$ can be covered with tiles. Therefore, if $n = 3k$, each of the three squares and the entire figure can be tiled.

We prove that $\Phi$ can be covered for any $n$ of the form $3k+1$. Put two horizontal tiles so that each of them closes two cells in the central and one in the side square. Then each of the side squares can be divided into rectangles $(3k+1) \times 3k$ and $3k \times 1$ each of which can be tiled. The unoccupied part of the central square can be divided into rectangles of sizes $3k \times 2$, $(3k+1) \times (3k-3)$ and $2 \times 3k$, each of which can be tiled.

Suppose that for some $n = 3k+2$ the figure $\Phi$ was covered with tiles. Since each square contains $n^2 = 3(3k^2 + 4k + 1) + 1$ cells, and two adjacent squares can have only one tile in common (intersecting both of them), the tiles entirely contained in the upper square cover it all, except for the right lower cell only. Paint the cells of the upper square in three colors as shown in the figure. It is easy to see that each tile covers one square of each color so the number of cells is the same for each color. But the number of cells of color $2$ is one more than the number of cells of color $1$, a contradiction.