If the numbers are $1, 1, 1, 1, -2$, (or $1, 1, 1, -2, -2$) there are $6$ sums that are equal to $0$ without the numbers themselves being $0$. Therefore, knowing $6$ sums (or less) to be $0$ is not enough to conclude that the numbers are all $0$.

Now we prove that $7$ is enough. Suppose we are given that $7$ sums are equal to $0$. These sums contain, in total, $21$ terms, while there are only $5$ numbers. By the Pigeonhole Principle it follows that there is a number that appears (at least) $5$ times. Suppose $x_1$ appears (at least) $5$ times. $x_1$ is involved in $6$ sums, so there is only one sum, say $x_1 + x_4 + x_5$, that is not necessarily $0$. All the other ones are $0$: $x_1 + x_2 + x_3 = x_1 + x_2 + x_4 = x_1 + x_2 + x_5 = x_1 + x_3 + x_4 = x_1 + x_3 + x_5 = 0$. The first three equations simplify to $x_3 = x_4 = x_5$, and then comparing the first and the last equation gives $x_2 = x_3 = x_4 = x_5$.

Finally, $x_1$ only participates in at most $6$ sums; the seventh one gives $x_2 = x_3 = x_4 = x_5 = 0$, and it follows immediately that $x_1 = x_2 = x_3 = x_4 = x_5 = 0$.