I also can't do the entire problem, but I can handle a more general case than the other answer: Barbara wins if Alan plays only 0s on a board of the form (4n)x(4n). (This is more general because it considers all possible ways of having 0s force a determinant to be zero, not just rows and columns. Admittedly, it's less general because it requires more of the size of the matrix.)
First, consider the 4x4 case. Label the matrix as follows:
aacd
bbcd
efgg
efhh
This pairs up the entries of the matrix. Then, Barbara's strategy is fairly easy: play in the matching pair of wherever Alan plays. One can check that no matter how Alan plays, there will be four entries of Barbara's whose product contributes to the overall determinant. If Barbara plays algebraically independent entries, then this implies the entire determinant is nonzero.
Extending this to (4n)x(4n) is fairly easy. Make n 4x4 blocks along the diagonal of the big matrix. If Alan plays in one of them, Barbara plays by the above strategy locally. If he plays in an off-diagonal block, Barbara simply helps Alan by playing 0 in that block. The end result will be a block diagonal matrix with nonzero determinant.
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As quid points out, this is more complicated than I originally made it seem.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 $5=a$.
Best Answer
0 always wins (regardless of 1's strategy). This is because any configuration with (a) 0s all in one row or (b) 0s all in one column or (c) 0s all in a 2x2 minor will win.
For the case 1 goes first, let 0 play in the center square. After which up to symmetry there are only four cases that need to be considered after 4 moves
(this being 0's strategy). It is easy to see that either 1 allows 0 to make a line in the 6th move, or 1 allows a block of the form
to form after the 6th move which guarantees win for 0.
When 0 goes first she wins faster.