This question is now also on http://cstheory.stackexchange.com/questions/4081/what-is-the-pathwidth-of-the-3d-grid-mesh-or-lattice-with-sidelength-k, where a discussion started, and one reference seems to be found.

Here, by 3D-grid of sidelength $k$ I mean the graph $G=(V,E)$ with $V= \{1,\ldots,k\}^3$ and $E=\{( (a,b,c) ,(x,y,z) ) \mid |a-x|+|b-y|+|c-z|=1 \}$, i.e., the nodes are placed at 3-dimensional integer coordinates between 1 and $k$, and a node is connected to the at most 6 other nodes that differ in precisely one coordinate by one.

What is the name of this graph? I'll use 3D grid, but perhaps 3D mesh or 3D lattice are what other people are used to.

What is the treewidth or pathwidth of this graph? Is this already published somewhere?

I know already that $tw(G) = 3/4 k^2 + O(k)$, i.e. it is smaller than $k^2$. To see this, we consider a path decomposition that "sweeps" the grid using mainly node-sets of the form $S_c= \{(x,y,z)\mid x+y+z = c\}$. Observe $|S_c| \leq 3/4 k^2 + O(k)$, $S_{3/2 k}$ being the largest such set. The sets between $S_c$ and $S_{c+1}$ are created by sweeping with a line and need $O(k)$ additional nodes to be separators. More precisely, use the sets $S_{c,d} = \{(x,y,z)\mid (x+y+z = c \wedge x \leq d ) \vee (x+y+z = c \wedge x \geq d ) \}$ as a path decomposition of $G$.

I also have an idea for a proof that shows $tw(G) = \Omega(k^2)$, but that is not finished yet.