Bob wins. A winning strategy: If Alice plays $a$ in the entry $(i,j)$, then Bob plays $-a$ in an empty entry of the $i^{th}$ row. That works because $2016$ is an even number and because, if $A$ is the full matrix, then $Au=0$ when $u=[1,\cdots,1]^T$.

Another strategy for Bob. If Alice writes $a$ in the entry $(i,1)$, Bob writes $a$ in the entry $(i,2)$. If Alice writes $a$ in the entry $(i,2)$, Bob writes $a$ in the entry $(i,1)$. If Alice plays in the column $j$ with $j\geq 3$, Bob writes $0$ in any empty entry $(i,j)$ with $j\geq 3$. Bob can always do this because $2016$ is even. At the end of the game, the resulting matrix $A$ has two equal columns, hence $\det A=0$.

Gian, Bob can write $a$ in the entry $(i,2)$ only when this entry is empty.

Yes, of course. Why is that an issue with my solution?

Sorry for cross-posting, but I believe that many users in CM forum don't frequently check HSO forum, so let me advertise it here. Here is an extension of this game, with replacement that the one who makes the last move seeks for a non-zero determinant. I found it rather hard to deal with.

Gian, your method is correct. Perhaps it would be better to add: "when Alice plays, if the entry $(i,1)$ (resp. $(i,2)$) is empty, then the entry $(i,2)$ (resp. $(i,1)$) too".

What about this? Bob pairs each entry of the form $(i,2k-1)$ with $(i,2k)$, for $i=1,\ldots ,2016$ and $k=1,\ldots ,1008$. If Alice writes $a$ in an entry, then Bob writes $a$ in the other entry of the pair. An easy induction shows that before each move of Alice, every pair has an even number of empty entries, and hence Bob can always make a move. When the matrix is complete, the columns $2k-1$ and $2k$ are equal for $k=1,\ldots ,1008$, and hence the matrix has zero determinant. I hope it is better explained now (it isn't the same method, but I think it's easier to explain).