For $n = 1$, it's trivial that first player can't win since he can't make any move.
Also for $n = 2$, both players can make the moves so the board will be full (by placing two $1 \times 2$ tiles), hence the first player can't win too.
But for $n \geq 3$,
First player will win if $n$ is odd and second player will win if $n$ is even.
The strategy is simply to mirror (to the center point of the grid) the opponent's moves. For first player if $n$ is odd, he can simply put $1 \times 3$ tile in the middle at the beginning.
For more details:
By mirroring to the center point of the grid means the cell on up left is "pairing" to the cell on down right. The cell on $i$-th row from above and $j$-th column from left is pairing to the cell on $i$-th row from below and $j$-th column from right.
Take a note that the $1 \times 2$ and $1 \times 3$ tiles will not occupy two cells which are pairing to each other (except the first $1 \times 3$ one from first player when $n$ is odd). Small proof: The pairing is on the different row and column.
Initially, every two cells which are pairing to each other are both unoccupied or occupied.
By previous statement, if one put the unoccupied cells; the other one may simply put their pairs since they must be unoccupied too. By doing so, the integrity is still same: every two cells which are pairing to each other are both unoccupied or occupied.
Therefore, the player who copied the opponents moves will win the game.