Solution:

Group the sixteen circles into rings: the center, the inner ring of five, the middle ring of five, and the outer ring of five, holding $p,$ $q,$ $r,$ and $s$ coins in total, respectively. The center has $5$ neighbors (the inner ring); each inner student has $3$ (the center and two middle students); each middle student has $4$ (two inner and two outer); each outer student has $4$ (two middle and two outer). A student with $k$ neighbors sends $\frac{1}{k}$ of their coins to each neighbor.

Summing the trades over each ring (for example, the outer ring receives a quarter of each middle student's coins twice over, which totals $\frac{r}{2}$) gives $$p = \frac{q}{3}, \qquad q = p + \frac{r}{2}, \qquad r = \frac{2q}{3} + \frac{s}{2}, \qquad s = \frac{r}{2} + \frac{s}{2}.$$

The first equation gives $q = 3p,$ the second then gives $r = 2(q - p) = 4p,$ and the last gives $s = r = 4p.$ The total is $p + 3p + 4p + 4p = 12p = 3360,$ so the center student had $p = 280$ coins.