Higher dimensional generalization of: Any quadrilateral tiles the plane? 
Any (non-self-intersecting) quadrilateral tiles the plane.




 
 
(MathWorld image.)

Q. What is the strongest known generalization of this statement to higher dimensions?
I.e., $\mathbb{R}^d$ filling with combinatorial cuboids?
Is Michael Goldberg's 37-yr-old paper the latest in $\mathbb{R}^3$?

Goldberg, Michael. "On the space-filling hexahedra." 
  Geometriae Dedicata 6.1 (1977): 99-108.




 
 
(Snippet from Goldberg.)

 A: I just stumbled over this, maybe you find it useful: 
As Douglas Zare already discussed in his comments, a polytope tiling $\mathbb{R}^n$ must have Dehn invariant zero. This statement has appeared at least twice in the literature: 
Once here: H. Debrunner: Über Zerlegungsgleichheit von Pflasterpolyedern mit Würfeln. Arch. Math. 35 (1980) 583-587.
And once here:  J.C. Lagarias and D. Moews: Polytopes that fill $\mathbb{R}^n$ and scissors congruence. Discrete and computational geometry 13 (1995), 573-583
A similar problem for tilings by translations has been solved by H. Hadwiger: Mittelpunktspolyeder und translative Zerlegungsgleichheit. Math. Nachr. 8 (1952) 53-58. This was also reproduced by Lagarias and Moews in the paper mentioned above.
See also the acknowledgement of priority by Lagarias and Moews, which discusses some of the history, all the literature references are taken from there. 
I have not found papers discussing if triviality of Dehn invariants is sufficient for tiling (at least in dimensions 3 and 4, where Dehn invariant and volume completely characterize scissors congruence). (Maybe it is possible to use the scissors congruence to the cube together with the cube tiling to prove the converse and construct a tiling?)
