Solution:
The answer is $2^{15} \cdot 8 \cdot 17$

We note that each $k \in \{1,2,3, \ldots, 16\}$ belongs to $2^{15}$ subsets of $\{1,2,3, \ldots, 16\}$. We reason as follows: we can assign 0 or 1 to $k$ according to whether it is not or in a subset of $\{1,2,3, \ldots, 16\}$. As there are 2 choices for a fixed $k$, $k$ belongs to half of the total number of subsets, which is $2^{16}$. Hence the sum is
$$\sum v(X) = 2^{15}(1+2+3+\cdots+16) = 2^{15} \cdot 8 \cdot 17 = 4456448$$