Nico can always ensure $16$ cells. Suppose that there is a partition like in the statement so that no $16$ cells can be taken. Then this partition has at most $1$ rectangle $1 \times k$ for each $k = 8, 9, 10, 11, 12, 13$; at most $2$ such rectangles for $k = 6, 7$; at most $3$ such rectangles for $k = 4, 5$; at most $5$ such rectangles $1 \times 3$, at most $7$ rectangles $1 \times 2$ and at most $15$ unit cells $1 \times 1$. Consequently the total area does not exceed

$$(8+9+10+11+12+13)+2(6+7)+3(4+5)+5 \cdot 3+7 \cdot 2+15 \cdot 1=160.$$

This is false because the $13 \times 13$ square has area $13^2 = 169$. Therefore $16$ cells can be taken regardless of how Facu plays.

On the other hand $17$ cells are not always achievable. Here is an example. The first row is untouched; the next $9$ are cut as follows:

$$12+1,11+2,10+3,9+4,8+5,8+5,7+6,7+6,5+4+4.$$

Rows $11$ and $12$ are $3+3+3+3+1$ and $2+2+2+2+2+2+1$; row $13$ is cut into $13$ unit cells. In summary, the answer is $16$ cells.