Let $F$ be the set of all $n$-tuples $(A_1, \ldots, A_n)$ such that each $A_{i}$ is a subset of $\{1, 2, \ldots, 1998\}$.  Let $|A|$ denote the number of elements of the set $A$.  Find
\[ \sum_{(A_1, \ldots, A_n)\in F} |A_1\cup A_2\cup \cdots \cup A_n|  \]
1. **Define the problem and notation:**
   Let \( F \) be the set of all \( n \)-tuples \( (A_1, A_2, \ldots, A_n) \) such that each \( A_i \) is a subset of \( \{1, 2, \ldots, 1998\} \). We need to find the sum:
   \[
   \sum_{(A_1, \ldots, A_n) \in F} |A_1 \cup A_2 \cup \cdots \cup A_n|
   \]
   where \( |A| \) denotes the number of elements in the set \( A \).

2. **Calculate the expected value of \( |A_1 \cup A_2 \cup \cdots \cup A_n| \):**
   For each element \( k \) in \( \{1, 2, \ldots, 1998\} \), consider the probability that \( k \) is in at least one of the sets \( A_1, A_2, \ldots, A_n \). 

   - The probability that \( k \) is not in a specific set \( A_i \) is \( \frac{1}{2} \).
   - The probability that \( k \) is not in any of the \( n \) sets is \( \left(\frac{1}{2}\right)^n = \frac{1}{2^n} \).
   - Therefore, the probability that \( k \) is in at least one of the sets is \( 1 - \frac{1}{2^n} \).

3. **Sum over all elements:**
   Since there are 1998 elements, the expected number of elements in \( A_1 \cup A_2 \cup \cdots \cup A_n \) is:
   \[
   1998 \left(1 - \frac{1}{2^n}\right)
   \]

4. **Calculate the total sum over all \( n \)-tuples:**
   There are \( 2^{1998n} \) possible \( n \)-tuples in \( F \) because each of the 1998 elements can either be in or out of each of the \( n \) sets independently.

   Therefore, the total sum is:
   \[
   2^{1998n} \cdot 1998 \left(1 - \frac{1}{2^n}\right)
   \]

\(\boxed{2^{1998n} \cdot 1998 \cdot \left(1 - \frac{1}{2^n}\right)}\)