Solution:

Take a primitive direction vector $(a, b, c)$ for the line. Any nonzero component of absolute value $2$ or more limits the line to at most $5$ grid points, so every component is $0$ or $\pm 1.$ Lines parallel to a coordinate axis contain $10$ points, never $8.$ If exactly one component is $0,$ the line lies in one of the $30$ planes parallel to a face of the cube ($3$ orientations, $10$ positions), and within that $10 \times 10$ grid it is a diagonal of slope $\pm 1$ shifted off center; the shift by $2$ in either direction from each of the two main diagonals gives exactly $8$ points. That is $4$ lines per plane, and each lies in only one of the $30$ planes: $4 \cdot 30 = 120$ lines.

Otherwise the direction is one of the four space-diagonal directions $(1, \pm 1, \pm 1)$ up to sign; by symmetry, count lines parallel to $(1, 1, 1)$ and multiply by $4.$ Such a line $(d + t,\ e + t,\ f + t)$ meets the grid in $10 - (\max(d,e,f) - \min(d,e,f))$ points, so exactly $8$ points means $\max - \min = 2.$ Normalizing the base point so that $\min(d, e, f) = 1,$ we need $(d, e, f)$ with entries in $\{1, 2, 3\}$ using both $1$ and $3:$ there are $27 - 8 - 8 + 1 = 12$ of them, hence $12$ lines per direction and $4 \cdot 12 = 48$ in all.

The total is $120 + 48 = 168.$