First player wins for $n$ at least five. First turn, name $0$. They name a number, say $-a$. Choose two numbers $b$ and $c$ such that neither $b$, $c$, nor $b+c=a$. Then name $b$, forcing them to name $-b$, then $c$, forcing them to name $-c$, then $-b-c$, winning. You can always choose two such numbers, since each positive number is missed by one of the following triples: $1+2=3, 1+3=4, 1+4=5, 2+3=5$.
As quid points out, this is more complicated than I originally made it seem. If $c\neq a+b$ but $a+b$ is in the interval, then the second player can name $a+b$ in response to $c$ and win.
To avoid this, if $1 <a\leq n-2$, choose $b=1$ and $c=a+1$. Neither $1$, $a+1$, nor $a+2=a$ so this works.
If $a\geq n-1$, choose $b=2$ and $c=1$. Since $n\geq 5$, neither $1$, $2$, nor $3=a$ so this works, and $a+b=a+2>n$.
If $a=1$, choose $b=2$ and $c=3$, so $c=a+b$ and neither $2$, $3$, nor $5=a$.
Eleven moves suffice.
François Brunault commented that the maker can get two moves to start on some hexagonal lattice (a lattice generated by unit vectors with an angle of $60$ degrees). In fact, by the fourth move, the maker can get three moves to start in a hexagonal lattice, and can choose these to be the vertices of an equilateral triangle of side $1$ so that the breaker has not played in this lattice yet.
Proof: Let the maker's first play be the origin $A$. Rotate the coordinates after the breaker's first move so that this play be ignored. Let the maker's second move be $B = (2x,0)$ where $x$ is a transcendental smaller than $1$. There are $4$ hexagonal lattices $H_{A,C}, H_{A,D}, H_{B,C}, H_{B,D}$ containing one element of $\lbrace (0,0), (2x,0)\rbrace$ and one element of $\lbrace C=(x,\sqrt{1-x^2}),D=(x,-\sqrt{1-x^2}) \rbrace$, the points of distance $1$ from the maker's first two moves. These hexagonal lattices have the property that the breaker hasn't played in any of them yet, and their pairwise intersections are empty or a point in $\lbrace A,B,C,D\rbrace.$ So, either the second play of the breaker misses both $H_{A,C}$ and $H_{B,C}$ or both $H_{A,D}$ and $H_{B,D}$. By symmetry, we can assume the second play of the breaker misses $H_{A,C}$ and $H_{B,C}$. Let the third play of the maker be $C$. This gives the maker an unopposed pair of points in both $H_{A,C}$ and $H_{B,C}$. The third play of the breaker can only be in one of these lattices. On the fourth move of the maker, the maker can play in the other to make an unopposed equilateral triangle of side length $1$. $\blacksquare$
Next, even if the maker is constrained to play in this lattice, the maker can force $5$ in a row by move $11$. Note that if the maker has an "open $3$" of $3$ points in row with two open spaces on either side, $- - \circ \circ \circ - -$, then the breaker has to respond immediately either just to one side or the other, or else the maker can make $5$ in a row in $2$ more moves. To avoid an explosion of cases, we'll let the breaker play both sides of an open $3$. Perhaps without this, the maker could force $5$ in a row in fewer moves.
We'll show that whatever the breaker's fourth move is, the maker can still force $5$ in a row by the eleventh move. By symmetry, we can assume the breaker's fourth move is in a sixth of the lattice between two of the perpendicular bisectors of the triangle's sides. We'll consider two possible moves within this sixth individually, and then all others.
o o x
o
5
o o x
o
x
5
o o x
o
x
x
6 5
o o x
o
x
x x
6 5
o o x
o
x x
x x
6 7 5
o o x
o
x x
x x
x 6 7 5 x
o o x
o
x x
x x
x 6 7 5 x
o o x
8 o
x x
x x x
x 6 7 5 x
o o x
8 o
x x x
x x x
x 6 7 5 x
o o x
8 o 9
x x x
This last play makes two open $3$s, with $7$ and with $8$, so with this $4$th move by the breaker, the maker can construct $5$ in a row by move $11$.
In the next sequence, I'll show the breaker's response immediately, again letting the breaker block both sides of open $3$s.
o o
o x
x
o o
o x
5
x
x
x 6 o o x
o x
5
x
x x
x 6 o o x
7 o x
5
x x
x x x
x 6 o o x
7 o x
5 8
x x x
x x x
x 6 o o x
7 o x
9 5 8
x x x
Again, the maker's 9th move creates two open $3$s, through the $7$ and through the $5$ and $8$, so with that choice of $4$th move by the breaker, the maker can construct $5$ in a row by move $11$.
Next we let the breaker's $4$th move block every other possibility in that sixth of the lattice, which will cover all of those possibilities simultaneously.
x x x x x
o o . x x x x x
o . x x x x x
. . . x x x x x
. . . . x x x
x x x x x
5 o o . x x x x x
o . x x x x x
. . . x x x x x
. . . . x x x
This play technically does not make an open $3$ since only one space to the right is open. So, the breaker does not have to respond to either side immediately. The other possibility is $2$ to the left. We will let the breaker play in all $3$ positions simultaneously.
x x x x x
x x 5 o o x x x x x x
o . x x x x x
. . . x x x x x
. . . . x x x
x x x x x x
x x 5 o o x x x x x x
o . x x x x x
6 . . x x x x x
x . . . . x x x
x x x x x x x
x x 5 o o x x x x x x
7 o . x x x x x
6 . . x x x x x
x x . . . x x x
x x x x x x x x
x x 5 o o x x x x x x
7 o . x x x x x
8 6 . . x x x x x
x x x . . . x x x
x x x x x x x x
x x 5 o o x x x x x x
7 o . x x x x x
8 6 9 . x x x x x
x x x . . . x x x
Again, the $9$th move creates two open $3$s, so the maker can construct $5$ in a row by move $11$.
Best Answer
It's possible to give a complete theory of 3x3 misere "X-only" tic-tac-toe disjunctive sums by introducing the 18-element commutative monoid $Q$ given by the presentation
$Q = \langle\ a,b,c,d\ | \ a^2=1,\ b^3=b,\ b^2c=c,\ c^3=ac^2,\ b^2d=d,\ cd=ad,\ d^2=c^2 \rangle\ $.
Such a "disjunctive" game of 3x3 neutral tic-tac-toe is played not just with one tic-tac-toe board (as has been previously discussed in this thread), but more generally with an arbitrary (finite) number of such boards forming the start position. On a player's move, he or she selects a single one of the boards, and makes an X on it (a board that already has a three-in-a-row configuration of X's is considered out-of-play). Play ends when every board has a three-in-a-row configuration, and the player who completes the last three-in-a-row on the last available board is the loser.
The game analyzed already in this thread corresponds to play on a 3x3 single board.
The monoid Q arises as the misere quotient of the impartial game
G = 4 + {2+,0}
{2+,0} is the canonical form of the 3x3 single board start position, and "4" is the nim-heap of size 4, which also happens to occur as a position in this game. I'm using the notation of John Conway's On Numbers and Games, on page 141, Figure 32.
One way to think of Q is that it captures the misere analogue of the "nimbers" and "nim addition" that are used in normal play disjunctive impartial game analyses, localized to the play of this particular impartial game, neutral 3x3 tic-tac-toe.
I performed these calculations partly using Mathematica, and partly using Aaron N. Siegel's "MisereSolver" program.
See also
http://arxiv.org/abs/math/0501315
http://arxiv.org/abs/math/0609825
http://arxiv.org/abs/0705.2404
http://www.miseregames.org
It's possible to build a dictionary that assigns an element of Q to each of the conceivable 102 non-isomorphic positions in 3x3 single-board neutral tic-tac-toe. (I mean "non-isomorphic" under a reflection or rotation of the board. In making this count, I'm including positions that couldn't be reached in actuality because they have too many completed rows of X's, but that doesn't matter since all those elements are assigned the identity element of Q). To determine the outcome of a multi-board position (ie, whether the position is an N-position -- a Next player to move wins in best play, or alternatively, a P-position-- second player to move wins), what a person does is multiply the corresponding elements of Q from the dictionary together, and reduce them via the relations in the presentation Q that I started with above, arriving at a word in the alphabet a,b,c,d.
If that word ends up being one of the four words {a, b^2, bc, c^2 }, the position is P-position; otherwise, it's an N-position.
I'm guessing the the 4x4 game does not have a finite misere quotient, but I don't know for sure.
If people want more details, I'm happy to send them. Google my name for my email address.
Best wishes
Thane Plambeck
Postscript (added 8 Jan 2013) Here's a paper http://arxiv.org/abs/1301.1672 I just put up in the arXiv that has more details.