Actually the answer is exactly as you suggested it to be $(-1+-1+-1)^n = (-1)^n 3^n$, and not 1.
Table[Sum[ Multinomial[n1, n2, n3] Boole[n1 + n2 + n3 == n] (-1)^( n1 - n2 + n3), {n1, 0, n}, {n2, 0, n}, {n3, 0, n}], {n, 0, 10}] {1, -3, 9, -27, 81, -243, 729, -2187, 6561, -19683, 59049}
You can easily check this with $n=1$. There are 3 possibly combination $n_1 = n_2 = 0 \land n_3 = 1$, or $n_1=n_3=0 \land n_2 =1$ or $n_2=n_3=0 \land n_1=1$.
$ \binom{1}{1,0,0}(-1) + \binom{1}{0,1,0}(-1) + \binom{1}{0,0,1}(-1) = (-1)+(-1)+(-1) = -3 \not= 1 $