Nice problem. The answer is (I think!) $\ln(2) \approx 0.693147181$.
Let's suppose that the correct placement of the 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.
Not shown: That $\delta$ is maximised by taking the points to infinity and then placing them at the points of a square. But that is all reasonable - When the points are close together, $\delta$ is small (comparable to the diameter).