On each non-boundary unit segment of an $8\times 8$ chessboard, we write the number of dissections of the board into dominoes in which this segment lies on the border of a domino. What is the last digit of the sum of all the written numbers?
1. **Define the problem and notation:**
   - Let \( N \) be the total number of ways to tile an \( 8 \times 8 \) chessboard with dominoes.
   - Consider each non-boundary unit segment on the chessboard. There are \( 56 \) horizontal and \( 56 \) vertical non-boundary unit segments, making a total of \( 112 \) such segments.

2. **Determine the number written on each segment:**
   - For any non-boundary unit segment, the number written on it is the number of tilings in which this segment lies on the border of a domino.
   - Let \( x_i \) be the number of tilings that include a domino across the \( i \)-th segment. Then, the number written on the \( i \)-th segment is \( N - x_i \).

3. **Sum the numbers written on all segments:**
   - The total sum of the numbers written on all \( 112 \) segments is:
     \[
     \sum_{i=1}^{112} (N - x_i) = 112N - \sum_{i=1}^{112} x_i
     \]

4. **Analyze the sum of \( x_i \):**
   - Notice that each tiling of the chessboard contributes exactly once to the count of \( x_i \) for each segment it crosses.
   - Since each tiling covers the entire board with \( 32 \) dominoes, and each domino covers exactly \( 2 \) segments, each tiling contributes to \( 64 \) segments in total.
   - Therefore, the sum \( \sum_{i=1}^{112} x_i \) counts each tiling \( 32 \) times:
     \[
     \sum_{i=1}^{112} x_i = 32N
     \]

5. **Substitute and simplify:**
   - Substitute \( \sum_{i=1}^{112} x_i = 32N \) into the total sum expression:
     \[
     112N - \sum_{i=1}^{112} x_i = 112N - 32N = 80N
     \]

6. **Determine the last digit:**
   - The last digit of \( 80N \) is determined by the last digit of \( 80 \times N \).
   - Since \( 80 \) ends in \( 0 \), the product \( 80N \) will always end in \( 0 \), regardless of \( N \).

Conclusion:
\[
\boxed{0}
\]