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It does not exceed ${n\choose \lceil n/2\rceil}$ by Sperner's theorem (the subsets with the same sum are not contained one in another). On the other hand, if $n$ is even and we get $n$ copies of 2, we get exactly ${n\choose n/2}$ subsets with sum $n$, so the bound is tight. If $n$ is odd, we may take 1 and $(n-1)$ copies of 2 to get ${n-1\choose (n-1)/2}$ subsets with sum $n$, which is worse than the above upper bound approximately by a factor of 2.