Solution:

We perform casework on the point three vertices away from $(0,0)$. By inspection, that point can be $( \pm 8, \pm 3),( \pm 7, \pm 2),( \pm 4, \pm 3),( \pm 3, \pm 2),( \pm 2, \pm 1)$ or their reflections across the line $y=x$. The cases are as follows:

If the third vertex is at any of $( \pm 8, \pm 3)$ or $( \pm 3, \pm 8)$, then there are 7 possible hexagons. There are 8 points of this form, contributing 56 hexagons.

If the third vertex is at any of $( \pm 7, \pm 2)$ or $( \pm 2, \pm 7)$, there are 6 possible hexagons, contributing 48 hexagons.

If the third vertex is at any of $( \pm 4, \pm 3)$ or $( \pm 3, \pm 4)$, there are again 6 possible hexagons, contributing 48 more hexagons.

If the third vertex is at any of $( \pm 3, \pm 2)$ or $( \pm 2, \pm 3)$, then there are again 6 possible hexagons, contributing 48 more hexagons.

Finally, if the third vertex is at any of $( \pm 2, \pm 1)$, then there are 2 possible hexagons only, contributing 16 hexagons.

Adding up, we get our answer of 216.