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EDIT: I was able to satisfy myself that the crossing probability is exactly $\frac14$, through a fairly (and probably unnecessarily) involved process. At most stages one can reduce to simpler calculations using symmetries or nice facts such as the probability-preserving projection map from a $(d-1)$-sphere in $\mathbb{R}^d$ onto the ball of its first $d-2$ components. For the final step I did have to compute an integral, though—the same one that tells you the angular momentum of a spinning coin (whose axis of rotation is in the plane of the coin).

I have started to suspect that in a certain precise sense an integral is unavoidable—that the boundary of crossing configurations is curved in ways that prevent abutment, just as the region north of $30^\circ$ latitude carries exactly $\frac14$ of the surface area of a perfect globe, but no finite collection of $4S$ rotated copies of it can be an exact $S$-fold cover. (The transcendental answer to the usual Sylvester's four-point problem in the disc also discourages the finite cover approach.)

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This is not really an answer but an over-long comment following up suggestions of Ian Agol and Bill Thurston.

Experiment suggests (with 97% confidence) that the crossing probability (in a specified or random projection, for two line segments with the four endpoints chosen independently uniformly at random with respect to Haar measure) is greater than $.2499$ and less than $.2501$. I have a motto never to compute by integrating what can be computed by symmetry, so the hope would be, for some integer $S$, to write a total of $4S$ symmetry-related expressions for the probability whose sum is identically $S$. So far any such trick eludes me; does anyone else see a way?

It would surprise me, for very large $n$, if even a constant factor improvement over a randomly chosen obvious projection is possible. I would wager, then (although I would hate to have to prove it) that $$\lim\limits_{n\rightarrow\infty} \frac{\bar{c}(n)}{n^2} = \frac18,$$ where $\bar{c}(n)$ is the expected value of crossing number for $n$ random points on the sphere.

The suggestion of a Voronoi-like spine whose dual would triangulate the knot complement is an interesting one. The individual faces are sections of hyperbolic paraboloid. It seems reasonable that their number would be strictly between linear and quadratic, although I don't yet see a good heuristic for guessing the correct order.