The dot dimension

Ways of choose three dots $\binom{17}{3}$

Ways of choose three dots that form a degenerate triangle horizontally $\binom{9}{3}$

Ways of choose three dots that form a degenerate triangle vertically $\binom{9}{3}$

Number of non-degenerate triangles $\binom{17}{3}-\binom{9}{3}-\binom{9}{3}=\boxed{512}$

Split the problem into two cases:

1. There is no middle point

    - There are 2 choices for one axis out of the two
    - There are 8C2 = 28 choices for two point on this axis
    - There are 8 choices for the final point on the the other axis
   

   There are $2 \times 28 \times 8 = 448$ choices

2. There is a middle point

       There are 8 choices for one point on the horizontal axis and 8 choices for one point on the vertical axis.
   

   There are $8\times8 = 64$ choices.

Adding these cases, you get $448 + 64 = 512$ total triangles.

Note that exactly $2$ points on our triangles must lie on the same axis. Any more, and the triangle would be degenerate. The number of ways to form a non-degenerate triangle such that $2$ points lie on the horizontal axis is $8\times T_8 = 288$ where $T_8 = 36$ is the $8\text{th}$ triangular number. The same will be true for triangles with $2$ points on the vertical axis, but some of those are identical to those which we already counted; we cannot double count the repeat triangles, namely those with a vertex at the center dot. Therefore there are only $8\times T_7 = 224$ new triangles. Add the counts: $224 + 288 = \boxed{512}$

For triangles that contain the centre point: 8 possibilities on vertical x 8 on horizontal = 64

For triangles that do not contain the central point and have two points on vertical: 8c2 on vertical x 8 on horizontal = 28 x 8 = 224

Same for two on horizontal = 224

Total = 512