Three. (This question is worth 16 points)
Let $F$ be the set of all ordered $n$-tuples $\left(A_{1}, A_{2}, \cdots, A_{n}\right)$, where $A_{i}(1 \leqslant i \leqslant n)$ are subsets of the set $\{1,2,3, \cdots, 2002\}$. Let $|A|$ denote the number of elements in the set $A$. For all elements $\left(A_{1}, A_{2}, \cdots, A_{n}\right)$ in $F$, find the sum of $\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right|$, that is,
$$
\sum_{\left(A_{1}, A_{2} \cdots, A_{n}\right) \in F}\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right| .
$$
Three, $\because\{1,2, \cdots, 2002\}$ has $2^{2002}$ subsets,
$$
\therefore|F|=\left(2^{2002}\right)^{n}=2^{2020 n} \text {. }
$$

Let $a$ be a fixed element in $\{1,2, \cdots, 2002\}$. To ensure that $a$ is not in $A_{1} \cup A_{2} \cup \cdots \cup A_{n}$, each $A_{i}(1 \leqslant i \leqslant n)$ must not contain $a$. Since the number of subsets of $\{1,2, \cdots, 2002\}$ that do not contain $a$ is $2^{2001}$, the number of ordered $n$-tuples $\left(A_{1}, A_{2}, \cdots, A_{n}\right)$ such that $A_{1} \cup A_{2} \cup \cdots \cup A_{n}$ does not contain $a$ is $\left(2^{2001}\right)^{n}=2^{2001 n}$. Therefore, there are $2^{2002 n}-2^{2001 n}$ ordered $n$-tuples $\left(A_{1}, A_{2}, \cdots, A_{n}\right)$ such that $a \in A_{1} \cup A_{2} \cup \cdots \cup A_{n}$.

Thus, when calculating the sum $\sum_{\left(A_{1}, A_{2}, \cdots, A_{n}\right) \in F}\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right|$, the element $a$ is counted $\left(2^{2002 n}-2^{2001 n}\right)$ times.
Since $\{1,2, \cdots, 2002\}$ has 2002 elements, we have
$$
\begin{array}{l}
\sum_{\left(A_{1}, A_{2}, \cdots, A_{n}\right) \in F} \mid A_{1} \cup A_{2} \cup \cdots \cup A_{n} \mid \\
=2002\left(2^{2002 n}-2^{2001 n}\right) .
\end{array}
$$