Solution:

For any nonempty $S$ we must have $0 \in S$. Now if we draw a directed graph of dependencies among the non-zero elements, it creates a balanced binary tree where every leaf has depth $3$. In the diagram, if $a$ is a parent of $b$ it means that if $b \in S$, then $a$ must also be in $S$.

We wish to find the number of subsets of nodes such that every node in the set also has its parent in the set. We do this with recursion. Let $f(n)$ denote the number of such sets on a balanced binary tree of depth $n$. If the root vertex is not in the set, then the set must be empty. Otherwise, we can consider each subtree separately. This gives the recurrence $f(n) = f(n-1)^2 + 1$. We know $f(0) = 2$, so we can calculate $f(1) = 5$, $f(2) = 26$, $f(3) = 677$. We add $1$ at the end for the empty set. Hence our answer is $f(3) + 1 = 678$.