By induction:
You've checked the base case manually.
Suppose true for $n$. In the case $n+1$, we select nonzero $v \neq w$. $v$ differs from $w$ in at least one entry, say the $i$th entry. Consider the vectors $\tilde{v}, \tilde{w}$ that have their $i$th entry deleted:
\begin{align}
\tilde{v} &= (v_1, \dots, v_{i-1}, v_{i+1},\dots, v_{n+1})
\\
\tilde{w} &= (w_1, \dots, w_{i-1}, w_{i+1},\dots, w_{n+1})
\end{align}
Case1
Suppose $\tilde{v} \neq \tilde{w}$. Then the induction hypothesis applies and the element-wise product of $A_n \tilde{v}$ with $A_n \tilde{w}$ has $2^{n-2}$ entries that are 1.
$A_{n+1}$ can be created by forming the matrix
\begin{align}
\begin{pmatrix}
A_n \\
A_n
\end{pmatrix}
\end{align}
and then inserting an $i$-th column with $n$ entries that are 0 and $n$ entries that are 1. Convince yourself that this enumerates exactly all of the rows of $A_{n+1}$ (possibly permuted, but this won't change our conclusions).
For the $n$ rows where the $i$th entry of $A_{n+1}$ is 0, nothing has changed from the $n$ case when we calculate $A_n \tilde{v}$ and $A_n \tilde{w}$ - we will still get $2^{n-2}$ entries that are 1 in the element-wise product.
For the $n$ rows where the $i$th entry of $A_{n+1}$ is 1, we have to do a little more work. Since $v$ and $w$ differ in the $i$-th entry, without loss of generality, assume that $v_i = 0, w_i = 1$. Then we have for an arbitrary $j$th row of this part of $A_{n+1}$
\begin{align}
(A_n \tilde{v})_j &= 0 \implies (A_{n+1} v)_j = 0
\\
(A_n \tilde{v})_j &= 1 \implies (A_{n+1} v)_j = 1
\\
(A_n \tilde{w})_j &= 0 \implies (A_{n+1} w)_j = 1
\\
(A_n \tilde{w})_j &= 1 \implies (A_{n+1} w)_j = 0
\end{align}
Lemma: I claim that $2^{n-1}$ entries of $A_n \tilde{v}$ were 1.
Since we know that $2^{n-2}$ entries of $A_n \tilde{v}$ and $A_n \tilde{w}$ were simultaneously 1, this means that $2^{n-2}$ entries of $A_n \tilde{v}$ were 1 while $A_n \tilde{w}$ was 0. Then these are precisely the entries where $A_{n+1} v$ and $A_{n+1} w$ are simultaneously 1. This means that we have $2^{n-2}$ entries of the element-wise product that are 1, from the rows of $A_{n+1}$ where column $i$ is 1.
Altogether, if I can prove my Lemma, combining the count from the rows of $A_{n+1}$ where column $i$ is 0 and 1, we have $2^{n-2}+2^{n-2} = 2^{(n+1)-2}$ entries of the element-wise product that are 1, as desired.
Proof of Lemma:
$\tilde{v}$ has, say $k$ entries that are 1. Then we are interested in rows of $A_n$ with an odd number of 1's that coincide with the entries of $v$ that are 1. There are
\begin{align}
\begin{pmatrix}
k \\
1
\end{pmatrix}
\end{align}
$k$-tuples that have precisely a single 1 in them,
\begin{align}
\begin{pmatrix}
k \\
3
\end{pmatrix}
\end{align}
$k$-tuples that have precisely three 1's in them, etc. So there are
\begin{align}
2^{n-k}
\sum\limits_{j=0}^{2j+1 \leq k}
\begin{pmatrix}
k \\
2j+1
\end{pmatrix}
=
2^{n-1}
\end{align}
rows in $A_n$that produce an odd number when multiplied by $\tilde{v}$. The factor $2^{n-k}$ appeared because $2^{n-k}$ entries of $v$ are zero and so having a 0 or a 1 in the corresponding column of a row of $A_n$ doesn't matter, meaning that both situations are counted.
Case 2
Suppose $\tilde{v} = \tilde{w}$. Then $A_n \tilde{v} = A_n \tilde{w}$ and by the above lemma, $2^{n-1}$ entries of the element-wise product are 1. When we construct $A_{n+1}$ as in Case 1 above, again without loss of generality, $v_i = 0, w_i = 1$, because they must differ in this entry.
All the rows where $A_{n+1}$ has a 0 in the $i$th column reduce to the $A_n$ case and so we have $2^{n-1}$ entries of the element-wise product that are 1.
All the rows where $A_{n+1}$ has a 1 in the $i$th column cannot have $A_{n+1} v = A_{n+1} w$ (because all entries of $v$ and $w$ are equal, except the $i$th) and so these rows cannot give 1's in the element-wise product.
This gives the total count of $2^{(n+1)-2}$ 1's in the element-wise product, as desired.
By the way, I believe your question is equivalent to the statement
Let $X$ be a set with $|X| = n$. Let $A,B \subseteq X$. Then there are
$2^{n-2}$ sets $Y\subseteq X$ such that $|A \cap Y|$ and $|B \cap Y|$
are odd.
Perhaps someone can find a shorter or more elegant proof using this observation.
