1. We need to count the number of functions $f: A \rightarrow A$ such that $f(1) < f(2) < f(3)$.
2. First, we choose 3 distinct values from the set $A = \{1, 2, 3, 4, 5, 6, 7, 8\}$ for $f(1), f(2),$ and $f(3)$. The number of ways to choose 3 values out of 8 is given by the binomial coefficient:
$$\binom{8}{3}$$
3. Once we have chosen 3 values, there is exactly one way to assign these values to $f(1), f(2),$ and $f(3)$ such that $f(1) < f(2) < f(3)$.
4. For the remaining 5 elements $f(4), f(5), f(6), f(7),$ and $f(8)$, each can be any of the 8 elements in $A$. Therefore, there are $8^5$ ways to assign values to these 5 elements.
5. Combining these, the total number of functions $f$ is:
$$\binom{8}{3} \times 8^5$$

The final answer is $\boxed{ \binom{8}{3} \times 8^5 }$