Orienting Simplices

If I think of these as objects in a 2-category, and I think of the $x$-direction of $\R^2$ as representing 'the flow of time as far as the 1-cells are concerned', and the $y$-direction as 'the flow of time as far as the 2-cells are concerned', then there's two possible sorts of 2-cell that these points (together with the lines and triangle that lie between them) can represent. One where two morphisms 'compose' across the 2-cell to yield a third, and one where one 'decomposes' into two morphisms.

But there is also a geometric sense in which we can take any cell and ask whether it looks like it's in the domain or codomain of a higher dimensional cell. For example, to ask whether $h$ is in the codomain of $\alpha$, we ask whether \[\det\left(\begin{matrix}a_x & a_y & 1 \\ c_x & c_y & 1 \\ b_x & b_y & 1 \end{matrix}\right) < 0 \] at least assuming the points are laid out as depicted, with $c_x - a_x$.

In general, if we have points $q, p^1, \ldots, p^n$ in $\R^{n}$ then \[\det\left(\begin{matrix} p^1_1 & p^1_2 & \cdots & p^1_{n} & 1\\ p^2_1 & p^2_2 & \cdots & p^2_{n} & 1\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ p^n_1 & p^n_2 & \cdots & p^n_{n} & 1\\ q_1 & q_2 & \cdots & q_{n} & 1 \end{matrix}\right) \over \det\left(\begin{matrix} p^1_1 & p^1_2 & \cdots & p^1_{n-1} & 1\\ p^2_1 & p^2_2 & \cdots & p^2_{n-1} & 1\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ p^n_1 & p^n_2 & \cdots & p^n_{n-1} & 1 \end{matrix}\right) \] is how far $q$ is — in the $n$-coordinate — from the hyperplane determined by $p^1, \ldots, p^n$. Thus, this will be positive if it's on the `positive $n$-coordinate side' of the hyperplane, and negative for the other.

So if we have a large sequence of points $S = (S^1, \ldots, S^m) \in (\R^\infty)^m$, then we can extract some coarse information about how they are shaped, as some data of type \[\Pi n < m. \Pi K : \mathcal P_n(S). \Pi q \in (S \setminus K) . \{\dom, \cod\} \] That is, for any $n < m$ and any $n$-sized subset $p^1, \ldots p^n \subseteq S$ and some $q \in S \setminus p^1, \ldots, p^n $ not already in that subset, we can project each of the $p^1, \ldots, p^n, q$ from $\R^\infty$ down to just $\R^n$ by taking their first $n$ coordinates, and ask whether the sign of the expression is above is positive or negative. We interpret this as telling us whether the $(n-1)$-cell determined by $p^1, \ldots p^n$ is in the domain or codomain of the $n$-cell determined by $q, p^1, \ldots p^n$.

Question

I know $\phi(1) = 1$ and $\phi(2) = 2$ and $\phi(3) = 12$, (because of the two triangle diagrams above, plus a factor of $3!$ for permutations of merely in which order the three vertices appear in $S$) and I think $\phi(4) = 576 = 4! \cdot 2 \cdot 2 \cdot 6$, because of a choice of permutation of the 4 vertices, a choice of how to flip all the 2-cells, a choice of how to flip the 3-cell, and the fact that 6 of the following 8 diagrams