Define $S(\emptyset) = 0$. Let $T_0 = \{3, 6\}$, $T_1 = \{1, 4, 7\}$, $T_2 = \{2, 5, 8\}$. For $A \subseteq T$, let $A_0 = A \cap T_0$, $A_1 = A \cap T_1$, $A_2 = A \cap T_2$, then

$$S(A) = S(A_0) + S(A_1) + S(A_2) \equiv |A_1| - |A_2| \pmod{3},$$

So $3|S(A)$ if and only if $|A_1| \equiv |A_2| \pmod{3}$. It follows that
$$(|A_1|, |A_2|) = (0, 0), (0, 3), (3, 0), (3, 3), (1, 1), (2, 2).$$

The number of nonempty subsets $A$ so that $3|S(A)$ is
$$2^2 \binom{3}{0} \binom{3}{0} + \binom{3}{0} \binom{3}{3} + \binom{3}{3} \binom{3}{0} + \binom{3}{3} \binom{3}{3} \\ + \binom{3}{1} \binom{3}{1} + \binom{3}{2} \binom{3}{2} - 1 = 87.$$

If $3|S(A)$ and $5|S(A)$, then $15|S(A)$. Since $S(T) = 36$, so the value $S(A)$ is 15 or 30 (if $3|S(A)$ and $5|S(A)$).

Furthermore,
$$\begin{aligned} 15 &= 8+7 = 8+6+1 = 8+5+2 = 8+4+3 \\ &= 8+4+2+1 = 7+6+2 = 7+5+3 \\ &= 7+5+2+1 = 7+4+3+1 = 6+5+4 \\ &= 6+5+3+1 = 6+4+3+2 \\ &= 5+4+3+2+1, \end{aligned}$$
$$36 - 30 = 6 = 5 + 1 = 4 + 2 = 3 + 2 + 1.$$
So the number of $A$ such that $3|S(A)$, $5|S(A)$, and $A \neq \emptyset$ is 17.

The answer is $87 - 17 = 70$.