7. Let the set $S=\left\{2^{0}, 2^{1}, \cdots, 2^{10}\right\}$, then the sum of the absolute values of the differences of any two distinct elements in $S$ is equal to
7. 16398.

Let the sum of the absolute values of the differences of any two distinct elements in the set $\left\{2^{0}, 2^{1}, \cdots, 2^{n}\right\}$ be $T_{n}$. Therefore,
$$
\begin{aligned}
T_{n} & =T_{n-1}+\left(2^{n}-2^{n-1}\right)+\left(2^{n}-2^{n-2}\right)+\cdots+\left(2^{n}-2^{0}\right) \\
& =T_{n-1}+2^{n} n-\left(2^{n}-1\right) .
\end{aligned}
$$

From the recursive relation, we get
$$
\begin{aligned}
T_{n} & =\sum_{k=1}^{n}\left(2^{k} k-2^{k}+1\right)=\sum_{k=1}^{n} 2^{k} k-\sum_{k=1}^{n} 2^{k}+\sum_{k=1}^{n} 1 \\
& =\sum_{k=1}^{n}\left(\sum_{i=k}^{n} 2^{i}\right)-\left(2^{n+1}-2\right)+n=\sum_{k=1}^{n}\left(2^{n+1}-2^{k}\right)-\left(2^{n+1}-2\right)+n \\
& =2^{n+1} n-\left(2^{n+1}-2\right)-\left(2^{n+1}-2\right)+n=2^{n+1}(n-2)+n+4 .
\end{aligned}
$$

When $n=10$, $T_{10}=2^{14}+14=16398$.