Sum to zero of all triples in a list - n^2 running time
Your first pair of loops builds the list sums. The length of this list is \$O(n^2)\$ in the length of the input.

In your second set of loops for each element in s_integers you filter all of sums to find those elements which match it. The runtime of this filter process is linear in the length of sums, and hence quadratic in the length of the input. You perform this quadratic process once for each member of the input list; the search loop is therefore cubic in the length of the input.

Your algorithm as a whole is therefore \$O(n^3)\$ in the length of the input.

To make the algorithm as a whole quadratic, you need to make the second (match) stage linear in the length of sums. Natural approaches would include:


Using a dictionary or some other map structure to lookup elements in the original either s_integers or sums by magnitude, in constant time; the match stage is then linear in the length of the other list. (Probably fastest on real input; it might be slightly tricky to check that there are no pathological cases, depending on which list you do lookup on).
Sorting sums (\$O(n \log n)\$), then effectively 'merging' sums and s_integers, (\$O(n)\$ in the length of the larger). This is easiest to prove to be \$O(n^2)\$