In Chazelle, B.; Lee, D. T. 's paper Computing 36, 1-16(1986), theorem 3 in page 15, it states that the maximum enclosing circle finding algorithm takes $O(n^2)$ time cost.
I think the key is still the $O(n^2)$ intersection graph construction algorithm it mentions early, (or see Edelsbrunner, H. (1987), Algorithms in Combinatorial Geometry, chapter 7). Afterwards the maximun enclosing circle finding should be straightforward.
Apparently, this problem is equivalent to find the point covered by maximum number of given circles and it is easily to know only those mostly $2n^2$ points intersected by given n circles need to be considered as candidates. (This also leads a $O(n^2log(n))$ algorithm directly )
However, by utilizing the above mentioned $O(n^2)$ construction algorithm, it leads an $O(n^2)$ algorithm for this problem, too. Because the intersection graph constructed with vertices as intersection points and edges as arcs is an Euler planar graph. So one can just travel all the arcs through an Euler cycle and an order of arcs indexed by the indexes of circles it belongs to and information of whether any arc is a "leaving away arc" (curved backwards) or "entering into arc" (curved forwards) for the met-during-travling vertice which this arc is incident on will be recorded.
By Jordan's theorem an intersection vertice is enclosed by a circle only if it meets a "leaving away arc" belonging to that circle first or has an incident arc belonging to that circle. So after the whole travel, the maximum enclosing circle can be easily found. It is simliar to the case of deciding the covering times for points with ordered intervals along a straight line, (or i.e. the 1D version of this enclosing problem), except the order has already been given by the travel. While by Euler's formular $V + E - F = 2$ for a planar graph, the total number of arcs is linear with the number of vertices, and since one does not need to record related information again when traveling back to vertices already visited, by handshaking lemma, the total time cost will be $O(n^2)$.