For all $n>1$, the Sierpinski triangle has $120^\circ$ rotational symmetry and no center triangle. So players $2$ and $3$ can always make the same move as player $1$ just did, rotated by $1/3$ and $2/3$ of a full turn respectively, and force player $1$ to lose.
For $n=1$, player $2$ loses.
For $n=2$, player $1$ loses.
For $n>2$, players $2$ and $3$ can force player $1$ to lose.
The more interesting specific question, I think, is which other coalitions can force a win.
A. For which $n$, if any, can players $1$ and $3$ force player $2$ to lose?
B. For which $n$, if any, can players $1$ and $2$ force player $3$ to lose?
Players $1$ and $3$ can always force player $2$ to lose (case A). For their initial moves, they make a beeline for player $2$'s corner. This ensures that player $2$ can't penetrate either of their territories (i.e., thirds of the board), and at least one opponent will penetrate player $2$'s territory. Once this happens, player $2$ has at least one fewer reachable triangle than either opponent. They can then fill in their own territories, and player $2$ will be the first to run out of moves.
Similarly, players $1$ and $2$ can force player $3$ to lose (case B), again by smothering player $3$ in his own territory.