@JeffE Agreed, certainly. Nit: I think maybe you've got the formula negated-- as the internal angles $\gamma_{i}$ increase, the area should increase, rather than decrease, right? Equivalently, the area is $2\pi$ minus the sum of the "turning angles" $\pi-\gamma_{i}$. That makes it easier to generalize from a polygon to any closed curve on the sphere: the summation turns into an integral.

@RobertIsrael see my previous comment (sorry, clumsy here-- I don't know how to cc multiple people at once, or from an answer)

@GerhardPaseman I posted an answer showing the equipotential surface at the origin, which, I think, confirms your conjecture that escape is possible through only the hexagonal faces.

@GerryMyerson by "side-triple" I mean a subset of the set of n sides, of cardinality 3. If there are n sides, then there are (n choose 3) side triples, so a quadrilateral has (4 choose 3) = 4 of them, and a pentagon has (5 choose 3) = 10 of them. But if you pick a particular triangulation of the dual to work with, that means you're selecting a particular n-2 of those (2 of them for a quad, 3 of them for a pentagon).

I figured it out. Given P, let Q be the cyclic polygon whose vertices are the outward unit normals of the sides of P. Then the weights are proportional to the triangle areas in an arbitrary triangulation of Q, weighting the incenters of the corresponding side-triples of P. The result is magically independent of the triangulation chosen. I'll write up an answer with pictures soon, and maybe a proof of triangulation-independence.

Regarding "Finding the nearest orthogonal matrix to a given matrix", aka orthogonal Procrustes problem or Wahba's problem or rigid point set registration (all minor variations on the same problem), the textbook solution is, indeed, to compute the SVD $M = U \Sigma V^T$, and throw away $\Sigma$, producing the answer $U V^T$, the orthogonal matrix closest to $M$. (Assume $M$ has positive determinant, to sidestep a mess.) But I noticed that, in the 2 dimensional case, SVD isn't needed at all. If $M = [[a,b],[c,d]]$, then the answer is simply $[[a+d,b-c],[c-b,a+d]] / \sqrt{(a+d)^2+(b-c)^2}$.