We will show that the maximum number of distinct distances is $7$. First we prove that the number of distinct distances cannot be more than $7$, then we will show that there is a sequence of blocks with $7$ distances.

The possible distances between two blocks in the sequence are the numbers $1$ to $8$. Therefore, there can certainly be no more than $8$ distinct distances. We will show that there is always at least one distance that does not occur.

If in a sequence the distances $8$ and $7$ do not both occur, we are done. Therefore, suppose we have a sequence in which these two distances do both occur. The distance $8$ can only occur between the very first and the very last block, so these should have the same letter on them, say $A$. The distance $7$ can only occur between the first and the eighth (second last) block, or between the second and the last block. Because both outer blocks have an $A$, the second or eighth block must also have an $A$. Then the sequence of blocks is $AAxxxxxxA$ (or the other way around: $AxxxxxxAA$), where on the place of $x$ are blocks with a $B$ or $C$. Now we see that the distance $6$ cannot occur anymore: the distances between the blocks with $A$ are $1$, $7$, and $8$, and the distances between the blocks with $B$ and the blocks with $C$ are at most $5$. Also in this case, there is at least one distance that does not occur.

We conclude that there is always one of the possible distances $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ that does not occur. Hence, the number of distinct distances cannot be more than $7$.

An example of a sequence of blocks where $7$ distinct distances occur, is $ABBACCBA$, with distances $4$, $4$, $8$; $1$, $5$, $6$; $1$, $2$, $3$ (only the distance $7$ is missing). So the maximal number of distinct distances is equal to $7$. ☐