How to write sums of combinations of signed sums?

We can write $C_n$, for every non-negative integer $n$, as $$C_n = \sum_{(s_1, s_2, \dots, s_n) \in \{-1, 1\}^n}s_1s_2\cdots s_n \cdot f\Big(s_1x_1 + s_2x_2 + \dots + s_nx_n\Big),$$ where $\{-1, 1\}^n := \Big\{(s_1, s_2, \dots, s_n) \mid s_i \in \{-1, 1\} \text{ for all positive integers } i \leq n\Big\}$.

The idea is to just use the "variables" $s_i$ to act as the signs.

For example, $\{-1, 1\}^3$ is defined to be $$\{(\color{red}{-1}, \color{red}{-1}, \color{red}{-1}), (\color{red}{-1}, \color{red}{-1}, \color{green}{+1}), (\color{red}{-1}, \color{green}{+1}, \color{red}{-1}), (\color{red}{-1}, \color{green}{+1}, \color{green}{+1}), (\color{green}{+1}, \color{red}{-1}, \color{red}{-1}), (\color{green}{+1}, \color{red}{-1}, \color{green}{+1}), (\color{green}{+1}, \color{green}{+1}, \color{red}{-1}), (\color{green}{+1}, \color{green}{+1}, \color{green}{+1})\},$$ so we have \begin{align*} C_3 = & \: (\color{red}{-1})(\color{red}{-1})(\color{red}{-1}) \cdot f\Big((\color{red}{-1})x_1 + (\color{red}{-1})x_2 + (\color{red}{-1})x_3\Big) \:+ \\ & \: (\color{red}{-1})(\color{red}{-1})(\color{green}{+1}) \cdot f\Big((\color{red}{-1})x_1 + (\color{red}{-1})x_2 + (\color{green}{+1})x_3\Big) \:+ \\ & \: (\color{red}{-1})(\color{green}{+1})(\color{red}{-1}) \cdot f\Big((\color{red}{-1})x_1 + (\color{green}{+1})x_2 + (\color{red}{-1})x_3\Big) \:+ \\ & \: (\color{red}{-1})(\color{green}{+1})(\color{green}{+1}) \cdot f\Big((\color{red}{-1})x_1 + (\color{green}{+1})x_2 + (\color{green}{+1})x_3\Big) \:+ \\ & \: (\color{green}{+1})(\color{red}{-1})(\color{red}{-1}) \cdot f\Big((\color{green}{+1})x_1 + (\color{red}{-1})x_2 + (\color{red}{-1})x_3\Big) \:+ \\ & \: (\color{green}{+1})(\color{red}{-1})(\color{green}{+1}) \cdot f\Big((\color{green}{+1})x_1 + (\color{red}{-1})x_2 + (\color{green}{+1})x_3\Big) \:+ \\ & \: (\color{green}{+1})(\color{green}{+1})(\color{red}{-1}) \cdot f\Big((\color{green}{+1})x_1 + (\color{green}{+1})x_2 + (\color{red}{-1})x_3\Big) \:+ \\ & \: (\color{green}{+1})(\color{green}{+1})(\color{green}{+1}) \cdot f\Big((\color{green}{+1})x_1 + (\color{green}{+1})x_2 + (\color{green}{+1})x_3\Big). \end{align*}

Using $[n]:=\{1,2,\ldots,n\}$ we can write $C_n, n\geq 1$ as \begin{align*} \color{blue}{C_n=\sum_{S\subseteq [n]}(-1)^{\left|[n]\setminus S\right|}f\left(\sum_{j\in S}x_j-\sum_{j\in[n]\setminus S}x_j\right)}\tag{1} \end{align*}

Example: $n=3$: Following (1) we can write $C_3$ as \begin{align*} C_3&=(-1)^{\left|[3]\setminus[3]\right|}f(x_1+x_2+x_3)\\ &\qquad+(-1)^{\left|[3]\setminus\{1,2\}\right|}f(x_1+x_2-x_3) +(-1)^{\left|[3]\setminus\{1,3\}\right|}f(x_1+x_3-x_2)\\ &\qquad+(-1)^{\left|[3]\setminus\{2,3\}\right|}f(x_2+x_3-x_1)\\ &\qquad+(-1)^{\left|[3]\setminus\{1\}\right|}f(x_1-x_2-x_3) +(-1)^{\left|[3]\setminus\{2\}\right|}f(x_2-x_1-x_3)\\ &\qquad+(-1)^{\left|[3]\setminus\{3\}\right|}f(x_3-x_1-x_2)\\ &\qquad+(-1)^{\left|[3]\setminus\emptyset\right|}f(-x_1-x_2-x_3) \end{align*} in accordance with OP's stated $C_3$.