Nice problem.  The answer is (I think!) $\ln(2) \approx 0.693147181$.  

Here is the idea.  The most symmetric placement of four points at infinity is at the corners of an ideal square.  In the upper half plane model we place these at $0, 1, \infty, -1$.  We place identical horospheres at each of these points.  So these are cyclically tangent, and all have the same 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 $y = H$ 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.