The greatest possible value of $n$ is $\binom{17}{6} = 12376$.
We can identify each row as a 6-element subset of $S_{18} = \{1, 2, \dots, 18\}$, so that the elements correspond to the numbers of columns which contain 1. Then we need to maximize the number of such subsets so that no three of them form a partition of $S_{18}$, since forming a partition is the only case for their complement to have an empty intersection.
First we construct an example of 12376 subsets. We simply need to consider all 6-element subsets of $\{1, 2, \dots, 17\}$. Since in any three subsets, there are $6 \times 3 = 18$ elements, there must be an element belonging to at least two of the subsets by the pigeonhole principle. Therefore, the condition is satisfied. In this example, there are $\binom{17}{6} = 12376$ subsets.
Next we show that $n \le 12376$. Suppose $S_1, S_2, \dots, S_n$ are the 6-element subsets corresponding to the rows, and let $T_1, T_2, \dots, T_{m-n}$ be the unused 6-element subsets, where $m = \binom{18}{6}$. Let $N$ be the number of pairs $(S_i, T_j)$ such that $S_i$ and $T_j$ are disjoint.
On the one hand, for each $S_i$, the 12 elements not in $S_i$ can form $\frac{1}{2}\binom{12}{6}$ unordered pairs of 6-element subsets. At least one subset in each pair must be a subset $T_j$ since no three of the $S_k$'s can form a partition of $S_{18}$. As each $T_j$ disjoint with $S_i$ belongs to exactly one such pair, this gives at least $\frac{1}{2}\binom{12}{6}$ pairs containing $S_i$. Thus, $N \ge \frac{n}{2}\binom{12}{6}$.
On the other hand, for each $T_j$, there are only $\binom{12}{6}$ subsets of $S_{18}$ which are disjoint with $T_j$. Hence, $N \le (m-n)\binom{12}{6}$.
Combining these, we obtain $\frac{n}{2}\binom{12}{6} \le (m-n)\binom{12}{6}$, i.e. $n \le \frac{2m}{3} = 12376$. This completes the proof.