Answer: $2^{n(n-1)/2}$.
Let $N = 2^n$ and $v_2(a) = s$ if and only if $2^s \mid a$ and $2^{s+1} \nmid a$, for a positive integer $a$. Let $v_2(A) = \{v_2(a) \mid a \in A\}$ for the set $A$.
Claim: A set $A$ is good if and only if $v_2(A) = \{0, 1, \dots, n-1\}$.
First we show that sets of the form $A_n = \{2^i b_i \mid b_i \text{ odd}, 0 \le i \le n-1\}$ are good. For $n=1$, this is trivial. For $n \ge 1$, assume $A_n$ is good:
$$\{S(X) \pmod N \mid X \subseteq A_n\} = \{0, 1, \dots, N-1\} \pmod N,$$
and
$$\{S(X) \pmod{2N} \mid X \subseteq A_{n+1}\} = \{S(X), S(X) + N \pmod{2N} \mid X \subseteq A_n\}.$$
This implies
$$\{S(X) \pmod{2N} \mid X \subseteq A_{n+1}\} = \{0, 1, \dots, 2N-1\} \pmod{2N}.$$
Hence, $A_{n+1}$ is good.
Assume that $A = \{a_1, a_2, \dots, a_n\}$ is good, and $M = 2^N - 1$. Since $2^N \equiv 1 \pmod M$ and
$$\{S(X) \pmod N \mid X \subseteq A\} = \{0, 1, \dots, N-1\} \pmod N,$$
we have
$$\prod_{j=1}^{n} (2^{a_j} + 1) = \sum_{X \subseteq A} 2^{S(X)} \equiv \sum_{k=0}^{N-1} 2^k \equiv 0 \pmod M.$$
Note $M = \prod_{i=0}^{n-1} (2^{2^i} + 1)$. Therefore, for any $0 \le i \le n-1$, there exists $1 \le j \le n$ such that $d = \gcd(2^{2^i} + 1, 2^{a_j} + 1) \ne 1$. Thus $2^{2^i} \equiv 2^{a_j} \equiv -1 \pmod d$, and $2^{2^{i+1}} \equiv 2^{2^{a_j}} \equiv 1 \pmod d$, implying $2^{i+1} \mid 2^{a_j}$. Since $2^s \equiv 1 \pmod d$ where $s \le i+1$, $s > i$. Therefore, $s = i+1$, and $a_j = 2^i b_i$, where $b_i$ is odd. Thus, $\{0, 1, \dots, n-1\} \subseteq v_2(A)$, and $|v_2(A)| = n$.
Thus, the number of good sets such that all elements are less than $2^n$ is:
$$2^{n-1} \times 2^{n-2} \times \dots \times 2^0 = 2^{\frac{n(n-1)}{2}}.$$