how many boolean functions exist that satisfy the condition
If we did not have this constraint, there would initially be $2^{2^n}$ possible boolean functions on $n$ boolean variables. We have $2^n$ unique inputs that we can either map to a 1 or 0 for any of them.

Now for this problem, we need to map a subset of those $2^n$ inputs because the rest will be implicitly mapped. For instance if we map $f(x_1, \neg x_2, \neg x_3) = 0$, this will implicitly map $f(\neg x_1, x_2, x_3) = 1$. This means only a subset of the $2^n$ inputs can be uniquely mapped, the rest will be implicitly mapped as described. For this, we need to determine how many pairs of inputs exist such that we have: $$(\{\ell_1, \ell_2, \ldots \ell_n\}, \{\neg \ell_1, \neg \ell_2, \ldots, \neg \ell_n\})$$ where $\ell_i$ is a literal (positive or negative) containing variable $x_i$. Intuitively, we should know that each pair contains 2 sets of $n$ variables so there should be $2^n / 2 = 2^{n-1}$ pairs possible. Another way to think about it would be to hold one variable constant positive, let's say $x_1$. We then define a new function: $$f_{x_1}(x_2, x_3, \ldots, x_n) = \ldots$$
In this function we "hold" $x_1$ to be positive so that we're only ever setting one of the sets in these pairs that we have described, thus we only have $2^{2^{n-1}}$ possible functions $f_{x_1}$. With this, we can now define $f$ pretty easily:

$$f(x_1, \ell_2, \ell_3, \ldots \ell_n) = f_{x_1}(\ell_1, \ell_2, \ldots, \ell_n)$$

$$f(\neg x_1, \neg \ell_2, \neg \ell_3, \ldots \neg \ell_n) = \neg f_{x_1}(\ell_1, \ell_2, \ldots, \ell_n)$$

We basically use $x_1$ to say "we're looking at the positive instance of this input" vs "we're looking at the negative instance of this input". Then this configuration of $f_{x_1}$ will ensure they are opposite. Consider the following example on 3 variables:

$$\begin{array}{|c|c|c|c|}
\hline x_1 &amp; x_2 &amp; x_3 &amp; f_{x_1}(x_2, x_3) &amp; f_{x_1}(\neg x_2, \neg x_3) &amp; f(x_1, x_2, x_3) \\
\hline 0 &amp; 0 &amp; 0 &amp; 1 &amp; 0 &amp; 1\\
\hline 0 &amp; 0 &amp; 1 &amp; 1 &amp; 1 &amp; 0\\
\hline 0 &amp; 1 &amp; 0 &amp; 1 &amp; 1 &amp; 0\\
\hline 0 &amp; 1 &amp; 1 &amp; 0 &amp; 1 &amp; 0\\
\hline 1 &amp; 0 &amp; 0 &amp; 1 &amp; 0 &amp; 1\\
\hline 1 &amp; 0 &amp; 1 &amp; 1 &amp; 1 &amp; 1\\
\hline 1 &amp; 1 &amp; 0 &amp; 1 &amp; 1 &amp; 1\\
\hline 1 &amp; 1 &amp; 1 &amp; 0 &amp; 1 &amp; 0\\
\hline 
\end{array}$$

With this we can see $f_{x_1}$ is clearly a function over $n-1$ variables and thus there are $2^{2^{n-1}}$ possible functions that satisfy this.