Note: I already posted the $3$-dimensional version of this question on Mathematics Stack Exchange, but no answer has been received so far, so I hope that MathOverflow can be a more suitable place in the hope of obtaining a proof or disproof of the general version of the original conjecture, holding for any finite number of dimensions $k$.
Here is the general version of the problem:
Let $k \in \mathbb{N}^+$ be given, so that $G_k := \{0, 1, 2\}^k \subset \mathbb{R}^k$ denotes the $3\times 3 \times \cdots \times 3$ grid consisting of $3^k$ integer points.
We emit a laser beam (infinitesimally thin, traveling in affine Euclidean $k$-dimensional space in straight lines) that must visit all $3^k$ points of $G_k$. We are allowed to place exactly $\frac{3^k-3}{2}$ ideal mirrors (dimensionless, freely positioned in $\mathbb{R}^k$, even at non-integer points) to redirect the beam. Thus, the beam follows a polygonal trail of $\frac{3^k-1}{2}$ straight segments and does not need to return to the starting spot, but it is allowed to cross itself (even multiple times).
General Conjecture. It is possible to traverse all the points of $G_k$ by using $\frac{3^k-3}{2}$ mirrors only if the emitter is outside the open $k$-cube $(0,2)^k$.
Remark. Years ago, I constructively proved that, given any positive integer $k$, the $\frac{3^k-1}{2}$-segment trails are always optimal (i.e., we cannot join all the $3^k$ points with any polygonal trail consisting of fewer than $\frac{3^k-1}{2}$ line segments). For $k < 3$, the conjecture is trivial, while, if $k = 3$ is given, I have explicitly provided $13$-segment trails starting from every point of $G$ except $(1,1,1)$ (see Figures $3$–$5$ of Solving the $106$ years old $3^k$ points problem with the clockwise-algorithm).
Thus, the $3$-dimensional version of the General Conjecture states that it is possible to traverse all the points of $G_3$ by using $12$ mirrors only if the emitter is outside the open cube $(0,2)^3$.
Any proof or counterexample, at least for the $k = 3$ case, is welcome.
