If $E$ is some set $E=\{x_1,x_2,\cdots,x_n,\cdots\}$ and $f:E\times \cdots \times E\to\mathbb{R}$ is a real valued function on $E$, and we have the sum,
$$\sum_{x_1,x_2,\cdots,x_n\in E}f(x_1,x_2,\cdots,x_n)$$
How would the above sum be computed? I'm not sure what it means to sum over multiple elements in a set at the same time.
This notation means that the sum should be taken over all $ n$-subsets $ \{x_1,x_2,\dots,x_n\} $ in $ E $.
For example, $ \sum_{a,b \in E} ab $ is the sum of all products of pairs of elements from $ E $.
If you like, you can rewrite the sum in the question as $$ \frac{1}{n!} \sum_{x_1 \in E} \sum_{x_2 \in E\setminus\{x_1\}} \cdots \sum_{x_n \in E \setminus \{x_1,\dots,x_{n-1}\}} f(x_1,x_2,\dots,x_n). $$ (The $1/n! $ is to prevent counting $ n$-tuples more than once, and we have to play with the sum bounds to prevent adding any $ x_i$s to themselves.)
If you want to take the sum over each ordering of the subsets, then one would just remove the $ 1/n! $ factor - then each distinct ordering of each subset will be run through $ f $ as well. This is probably what you want if $ f $ isn't symmetric in its arguments.