Solution:
Answer: 376

Define a main plane to be one of the $xy$, $yz$, $zx$ planes. Define a space diagonal to be a set of collinear points not parallel to a main plane. We classify the lines as follows:

(a) Lines parallel to two axes (i.e. orthogonal to a main plane). Notice that given a plane of the form $v = k$, where $v \in \{x, y, z\}, k \in \{0,1,2,3\}$, there are 8 such lines, four in one direction and four in a perpendicular direction. There are $4 \times 3 = 12$ such planes. However, each line lies in two of these $(v, k)$ planes, so there are $\frac{8 \times 4 \times 3}{2} = 48$ such lines. Each of these lines has 4 points, so there are 4 possible ways to choose 3 collinear points, giving $4 \times 48 = 192$ triplets.

(b) Diagonal lines containing four points parallel to some main plane. Consider a plane of the form $(v, k)$, as defined above. These each have 2 diagonals that contain 4 collinear points. Each of these diagonals uniquely determines $v, k$ so these diagonals are each counted once. There are 12 possible $(v, k)$ pairs, yielding $12 \times 2 \times 4 = 96$ triplets.

(c) Diagonal lines containing three points parallel to some main plane. Again, consider a plane $(v, k)$. By inspection, there are four such lines and one way to choose the triplet of points for each of these lines. This yields $4 \times 12 = 48$ triplets.

(d) Main diagonals. There are four main diagonals, each with 4 collinear points, yielding $4 \times 4 = 16$ triplets.

(e) Space diagonals containing three points. Choose one of the points in the set $\{1,2\}^{3}$ to be the midpoint of the line. Since these 8 possibilities are symmetric, say we take the point $(1,1,1)$. There are four space diagonals passing through this point, but one is a main diagonal. So each of the 8 points has 3 such diagonals with 3 points each, yielding $8 \times 3 = 24$ ways.

Adding all these yields $192 + 96 + 48 + 16 + 24 = 376$.