Solution:

We approach the problem by finding the minimum number of triangles with a pair of edges of the same color; we call such triangles "isosceles". We can count the number of such triangles by instead considering:
The number of isosceles triangles is equal to the number of pairs of adjacent edges of the same color.
So, if we let $n_{i, v}$ denote the number of edges of the $i$th color touching vertex $v$, the number of isosceles triangles can be written as
$$\sum_{v \text{ vertex}} \sum_{i=1}^{8}\left(\begin{array}{c} n_{i, v} \\ 2 \end{array}\right)$$
Since $\sum_{i=1}^{8} n_{i, v}=16$, by Jensen's Inequality we actually have
$$\sum_{v \text{ vertex}} \sum_{i=1}^{8}\left(\begin{array}{c} n_{i, v} \\ 2 \end{array}\right) \geq \sum_{v \text{ vertex}} 8\left(\begin{array}{c} 16 / 8 \\ 2 \end{array}\right)=17 \cdot 8=136$$
So, the number of triangles with all different colors is at most $\left(\begin{array}{c}17 \\ 3\end{array}\right)-17 \cdot 8=544$. We leave the construction of a maximal example as an exercise to the reader.