Solution:

The number is $2^{n+1}$. We argue by complete induction on $n$ and set $B_{n}=\{1,2, \ldots, 2^{n}\}$.

Base case, $n=1$. This case is clear: all four subsets of $B_{1}=\{1,2\}$ are funny.

Inductive step, $n \longrightarrow n+1$. So suppose it is already known that $B_{n}$ has exactly $2^{n+1}$ funny subsets. For each funny set $A$ of $B_{n}$ construct the sets

$$C_{n}(A)=\{2^{n+1}-m \mid m \in B_{n} \backslash A,\ m<2^{n}\},$$

$$M_{n}(A)=A \cup C_{n}(A)$$

and

$$N_{n}(A)=M_{n}(A) \cup\{2^{n+1}\}.$$

Evidently $A$, $C_{n}(A)$ and $\{2^{n+1}\}$ are pairwise disjoint. The sets $M_{n}(A)$ and $N_{n}(A)$ are funny: to see this, let $x, y$ be distinct elements of $B_{n+1}$ whose sum is a power of two; without loss of generality let $x<y$.

In the case $1 \leq x<y \leq 2^{n}$, that is $x, y \in B_{n}$, exactly one of the two is contained in $M_{n}(A)$ resp. $N_{n}(A)$, since $A$ is funny.

In the case $1 \leq x \leq 2^{n}<y \leq 2^{n+1}$ we have $2^{n}<x+y<2^{n+2}$, hence $x+y=2^{n+1}$; since $x \geq 1$, we have $y \neq 2^{n+1}$. By the choice of $C_{n}(A)$, either $x$ lies in $A$ or $y$ lies in $C_{n}(A)$.

The case $2^{n}<x<y \leq 2^{n+1}$ cannot occur, since then $2^{n+1}<x+y<2^{n+2}$.

Conversely, if $M$ is a funny subset of $B_{n+1}$, then evidently $A:=M \cap B_{n}$ is a funny subset of $B_{n}$. The element $y=2^{n+1}-x$ of $B_{n+1}$ with $1 \leq x<2^{n}$ is contained in $M$ if and only if $x \in B_{n}$ is not contained in $A$, since $x+y=2^{n+1}$ is a power of two. Thus, knowing $M \cap B_{n}$, the funniness of $M$ determines, for all elements $y$ of $B_{n+1}$ with $2^{n}<y<2^{n+1}$, whether they are contained in $M$; the only remaining freedom is whether $2^{n+1}$ is an element. Hence $M$ equals either $M_{n}(A)$ or $N_{n}(A)$, and to each funny subset of $B_{n}$ exactly two funny subsets of $B_{n+1}$ can be assigned. This completes the proof.