Now, choosing a total order on the vertices seems to be a necessary and sufficient way for telling me which way the arrows go: they go from less to more, according to the order. Any acyclic graph with exactly one edge between every pair of vertices induces a unique total order on the vertices.
But I don't know what the higher-dimensional analogues are of this nice characterization of the 1-dimensional case.
Doing some manual calculations I figured that there are 1, 2, 12, 576 directed simplices of dimension 0, 1, 2, 3, if I label the vertices. In the 1-dimensional case, if the vertices are labelled a and b, I have a choice of whether the arrow runs a → b or b → a. In two dimensions, I have a choice of 3!=6 permutations of the three vertices that tell me how the arrows run. No matter what, I have a "long" path of two arrows joined end-to-end, and a "short" path of the remaining third arrow. Then I have a choice of whether the 2-cell goes from the long path to the short, or from the short path to the long. So that's a total of 6*2=12 distinct cells. In the 3-dimensional case, I thiiiiink I correctly figured there to be 4 types of 3-cell that have a codomain of a single face of the tetrahedron, (one for each face) 4 types of 3-cell that have a domain of a single face of the tetrahedron, (again one for each face) and 4 types that have a codomain and domain consisting of two complementary pairs of faces. (although there are 6 ways of choosing 2 faces of the 4, only 4 of the 6 seem to "compose well").
So that's (4 + 4 + 4), times 24 for the symmetry of the four vertices, times 2 for the choice of orientation of the 3-cell, which leaves 576.
Now of course we waltz over to OEIS and it tells us that 1,2,12,576 counts Latin Squares. What the fuuuuck.
Then I ask google for "latin squares directed simplices" to see if this is known and alls I get is this paper, which hints at some work done in 1902 connecting latin squares to simplices somehow, but does not answer my question.
As usual, I'm left shaking my head at how wacky math is.
Here's another maddening thing about the Latin square thing. If you look at http://oeis.org/A000315 the count of the reduced latin squares, the number of factors of 2 in the prime decomposition goes like: 2, 3, 6, 10, 17, 21, 28, 35 which totally looks like it's bounded below by the triangular numbers until you get to that 35 which is less than 36. The other anomalies being the 2 and the 17, but if there's some symmetry phenomenon that guarantees a redundancy factor of 2n(n-1)/2, it doesn't matter if you are still an even integer after that. What the hell.