<s>[INST] Are the following identities well known? $$
x \cdot y = \frac{1}{2 \cdot 2 !} \left( (x + y)^2 - (x - y)^2 \right)
$$
$$
\begin{eqnarray}
x \cdot y \cdot z &=& \frac{1}{2^2 \cdot 3 !} ((x + y + z)^3 - (x + y - z)^3 \nonumber \\
&-& (x - y + z)^3 + (x - y - z)^3 ), 
\end{eqnarray}
$$
$$
\begin{eqnarray}
x \cdot y \cdot z \cdot w &=& \frac{1}{2^3 \cdot 4 !} ( (x + y + z + w)^4 \nonumber \\
&-& (x + y + z - w)^4 - (x + y - z + w)^4 \nonumber \\
&+& (x + y - z - w)^4 - (x - y + z + w)^4 \nonumber \\
&+& (x - y + z - w)^4 + (x - y - z + w)^4 \nonumber \\
&-& (x - y - z - w)^4 ). 
\end{eqnarray}
$$
The identity that rewrites a product of $n$ variables $( n \ge 2$, $n \in \boldsymbol{\mathbb{Z}_+})$ as additions of $n$ th power functions is as given below:
$$
\begin{eqnarray}
& &x_0 \cdots x_1 \cdot x_{n-1} = \frac{1}{2^{n - 1} \cdot n !} \cdot \sum_{j = 0}^{2^{ n - 1} -1} ( - 1 )^{\sum_{m = 1}^{n - 1} \sigma_m(j)} \times ( x_0 + (-1)^{\sigma_1(j)} x_1 + \cdots + (- 1)^{\sigma_{n - 1}(j)} x_{n - 1})^n, \\
& &\sigma_m(j) = r(\left\lfloor \frac{j}{2^{m - 1}}\right\rfloor, 2), m (\ge 1), j (\ge 0) \in \mathbb{Z}_+, \; \left\lfloor x \right\rfloor = \max \{ n \in \mathbb{Z}_+ ; n \le x, x \in \mathbb{R} \}
\end{eqnarray}
$$
where $r(\alpha, \beta)$,  $\alpha, \beta (\ge 1) \in \mathbb{Z}_+$ means the remainder of the division of $\alpha$ by $\beta$ such that $r(\alpha, \beta)$ $=$ $\alpha - \beta \left\lfloor \frac{\alpha}{\beta}\right\rfloor$
 [/INST] Although not the exactly the same due to $2^{n-1}$ instead of $2^n$ terms, the OP's formula seems to be essentially the well-known polarization formula for homogeneous polynomials, which is stated as following:

Any polynomial $f$, homogeneous of degree $n$ can be written as $f(x)=H(x,\ldots,x)$ for a specific multilinear form $H$. One has the following polarization formula for $H$ (see also this MO post):
  \begin{equation*}
 H(x_1,\ldots,x_n) = \frac{1}{2^n n!}\sum_{s \in \{\pm 1\}^n}s_1\ldots s_n f\Bigl(\sum\nolimits_{j=1}^n s_jx_j\Bigr)
\end{equation*}

In your case, $f=x^n$, so $H(x_1,\ldots,x_n) = x_1\cdots x_n$ (please note off by 1 indexing).
</s>