The answer is $\ln(2) \approx 0.693147181$.  

Claim: the the correct placement of the four points at infinity is at the corners of an ideal square.  

With the claim in hand, we can compute $\delta$ in the upper half plane model.  We place the points at $0, 1, \infty, -1$.  We place identical horocircles at each of these points.  These are cyclically tangent, and all have the same minimal distance from the point $i$.  The points of tangency are cyclically permuted by the order four rotation about $i$.  If we take boundary of the horosphere about $\infty$ to be the line $y = H$ then we discover that the order four element (fixing $i$) sends $1 + iH$ to $-1 + 2i/H = -1 + iH$.  Thus $H = \sqrt{2}$.  So $\delta$ is twice the distance from $i$ to $i\sqrt{2}$ and we are done.

The proof of the claim appears to be difficult.  We have to prove that, given four material points, we can increase $\delta$ by first moving them "outward" to lie on a circle (tricky), then to lie symmetrically on the circle (medium), and then increase the radius of the circle to infinity (easy).