Optimal covering trails for every $k$-dimensional cubic lattice $\mathbb{N}^k := \{(x_1, x_2, \dots, x_n) : x_i \in \mathbb{N} \wedge n \geq 3\}$

Question

After a dozen years spent investigating this particular class of problems, I finally give up and I wish to ask you if any improvement is achievable from here on.

The general problem is as follows:
Let the $k$-dimensional grid $\mathcal{H}(n,k):=\{0,1,2,\ldots,n-1\} \times \cdots \times \{0,1,2,\ldots,n-1\}$ be given in the Euclidean space $\mathbb{R}^k$. How many line segments connected at their endpoints have the minimum-link covering trails that join all the points of $\mathcal{H}(n,k)$ (i.e., the optimal polygonal chain that visits all the given $n^k$ points by spending the minimum number of line segments)?

Then, let us call $h(n,k)$ the mentioned minimum number of links.
Accordingly, let us denote with $h_l(n,k)$ the proven lower bound for $h(n,k)$ and with $h_u(n,k)$ the corresponding upper bound so that we can state the following set of relations.

\begin{equation} h(n,1)=1, \forall n. \end{equation} \begin{equation} h(n,2)=2 \cdot (n-1), \forall n \geq 3. \end{equation} \begin{equation} h(1,k)=1, \forall k. \end{equation} \begin{equation} h(2,k)=3 \cdot 2^{k-2}, \forall k \geq 2. \end{equation} \begin{equation} h(3,k)=\frac{3^k-1}{2}, \forall k. \end{equation} \begin{equation} h(n,k) \leq \Bigg(\Bigl\lfloor{\frac{3}{2} \cdot n^2}\Bigl\rfloor - \Bigl\lfloor{\frac{n-1}{4}}\Bigl\rfloor + \Bigl\lfloor{\frac{n+1}{4}}\Bigl\rfloor - \Bigl\lfloor{\frac{n+2}{4}}\Bigl\rfloor + \Bigl\lfloor{\frac{n}{4}}\Bigr\rfloor + n - 1 \Bigg) \cdot n^{k-3} - 1, \forall n\geq 3 \wedge \forall k \geq 3 \end{equation} \begin{equation} h(n,k) \geq \frac{n^k-1}{n-1}, \forall n \geq 3 \wedge \forall k. \end{equation} \begin{equation} h(n,k) \leq (h_u(n,3)+1) \cdot n^{k-3}-1, \forall n\geq 3 \wedge \forall k \geq 3. \end{equation}

Thus, what I have currently found is resumed in the table below, where the strictly proven results are highlighted in the green cells, while the proven bounds are shown in the orange cells.

$\{0,1,2,\dots,n-1\}\times\{0,1,2,\dots,n-1\}\times \dots \times\{0,1,2,\dots,n-1\}$ points problem in $\mathbb{R}^k$." />

Conjecture: $h(n,k)=\frac{n^k-1}{n-1}+c \cdot (n-3)$, where $c = k-1$ iff $h(4,3)=23$, $c = 1$ iff $h(4,3)=22$, or even $c = 0$ iff $h(4,3)=21$ (see also this Previous discussion about finding the value of $h(4,3)$).

Any idea to go further (can we see some kind of pattern in the table)?
Thanks in advance!