[Solution] We prove a more general conclusion:
For any $n$ positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, the sum $\sum \frac{1}{Q(\sigma)}=$ $S\left(a_{1}, a_{2}, \cdots, a_{n}\right)=\frac{1}{a_{1} a_{2} \cdots a_{n}}$.

We use induction on $n$.
When $n=1$, $S\left(a_{1}\right)=\sum \frac{1}{Q(\sigma)}=\frac{1}{a_{1}}$, so the conclusion holds for $n=1$.
Assume the conclusion holds for $n=k$. We will prove it for $n=k+1$.
Indeed, the $(k+1)$ numbers $a_{1}, a_{2}, \cdots, a_{k}, a_{k+1}$ can be divided into $k+1$ classes, each containing $k!$ permutations that end with the same element $a_{i}$. The last factor of $Q(\sigma)$ for these permutations is $S_{k+1}(\sigma)=a_{1}+a_{2}+\cdots+a_{k}+a_{k+1}$. The first $k$ factors are given by all possible permutations of the other $k$ elements.

By the induction hypothesis, the sum of $\frac{1}{Q(\sigma)}$ for these $k!$ permutations is
$$\frac{a_{i}}{a_{1} a_{2} \cdots a_{k} a_{k+1} S_{k+1}(\sigma)}$$

Summing over $i$ gives
$$\begin{aligned} \sum \frac{1}{Q(\sigma)} & =S\left(a_{1}, a_{2}, \cdots, a_{k+1}\right) \\ & =\frac{1}{a_{1} a_{2} \cdots a_{k} a_{k+1} S_{k+1}(\sigma)} \cdot \sum_{i=1}^{k+1} a_{i} \\ & =\frac{1}{a_{1} a_{2} \cdots a_{k} a_{k+1}} \end{aligned}$$

By the principle of mathematical induction, the general conclusion holds.
Thus, the result for this problem is
$$\sum \frac{1}{Q(\sigma)}=\frac{1}{1 \cdot 2 \cdot 2^{2} \cdots 2^{n-1}}=2^{-\frac{n(n-1)}{2}}$$