Winning strategies in multidimensional tic-tac-toe
I will quote some results and problems from the book Combinatorial Games: Tic-Tac-Toe Theory by József Beck, some of which were also quoted in this answer.
The terms "win" and "draw" refer to the game as ordinarily played, i.e., the first player to complete a line wins. The term "Weak Win" refers to the corresponding Maker-Breaker game, where the first player ("Maker") wins if he completes a line, regardless of whether the second player has previously completed a line; in other words, the second player ("Breaker") can only defend by blocking the first player, he cannot "counterattack" by threatening to make his own line. (Note that ordinary $3\times3$ tic-tac-toe is a Weak Win.) A game is a "Strong Draw" if it is not a Weak Win, i.e., if the second player ("Breaker") can prevent the first player from completing a line.
Theorem 3.1 Ordinary $3^2$ Tic-Tac-Toe is a draw but not a Strong Draw.
Theorem 3.2 The $4^2$ game is a Strong Draw, but not a Pairing Strategy Draw (because the second player cannot force a draw by a single pairing strategy.)
Theorem 3.3 The $n\times n$ Tic-Tac-Toe is a Pairing Strategy Draw for every $n\ge5.$
For a discussion of Oren Patashnik's computer-assisted result that $4^3$ tic-tac-toe is a first player win, Beck refers to Patashnik's paper:
Oren Patashnik, Qubic: $4\times4\times4$ Tic-Tac-Toe, Mathematics Magazine 53 (1980), 202-216.
Not much more is known about multidimensional tic-tac-toe, as can be seen from the open problems:
Open Problem 3.2 Is it true that $5^3$ Tic-Tac-Toe is a draw game? Is it true that $5^4$ Tic-Tac-Toe is a first player win?
The conjecture that "if there is a winning strategy on $a^d$, there is one also on $a^{d'}$ for any $d'\geq d$" is given as an open problem:
Open Problem 5.2 Is it true that, if the $n^d$ Tic-Tac-Toe is a first player win, then the $n^D$ game, where $D\gt d$, is also a win?
Open Problem 5.3. Is it true that, if the $n^d$ game is a draw, then the $(n+1)^d$ game is also a draw?
To see that the intuition "adding more ways to win can't turn a winnable game into a draw game" is wrong, consider the following example of a tic-tac-toe-like game, attributed to Sujith Vijay: The board is the set $V=\{1,2,3,4,5,6,7,8,9\};\ $ the winning sets are $\{1,2,3\},$ $\{1,2,4\},$ $\{1,2,5\},$ $\{1,3,4\},$ $\{1,5,6\},$ $\{3,5,7\},$ $\{2,4,8\},$ $\{2,6,9\}$. As in tic-tac-toe, the two players take turns choosing (previously unchosen) elements of $V;$ the game is won by the first player to succeed in choosing all the elements of a winning set. It can be verified that this is a draw game, but the restriction to the board $\{1,2,3,4,5,6,7\}$ (with winning sets $\{1,2,3\},$ $\{1,2,4\},$ $\{1,2,5\},$ $\{1,3,4\},$ $\{1,5,6\},$ $\{3,5,7\}$) is a first-player win.
$4^3$ ("Qubic") is a win for the first player. According to this link, it was first proved by Oren Patashnik in 1980. The proof is complicated. It took 12 years for this proof to be converted into a practical computer algorithm; I was present at the 1992 Computer Olympiad where the program of Victor Allis and Patrick Schoo romped to victory.
Based on what I remember from playing such games twenty-five years ago,
The $3^3$ version is a guaranteed win for the first player, by going in the middle square. There are so many lines through it that the first player can always force moves. After the 2nd player places his irrelevant O, the first player chooses a plane through the middle X that doesn't contain that one O. Then, the game is henceforth played in that plane, where the first player effectively takes the first two moves in a row and forces a win.
Because of the forced moves, $3^n$ is equally a win for the first player for all $n\ge 3$.
The $4^3$ version does not admit such a simple strategy (nor does the $4^4$).