Answer: 19

Solution:

Track the three gaps between consecutive frogs around the circle; they start at $(4, 4, 4)$ and always sum to $12.$ If the frogs jump by $X_1, X_2, X_3 \in \{\pm 1\},$ the gaps change by $X_2 - X_1,$ $X_3 - X_2,$ $X_1 - X_3,$ so each gap stays even and the process stops exactly when some gap becomes $0.$ Enumerating the $8$ equally likely sign choices: from $\{4,4,4\},$ the state stays with probability $\frac{2}{8}$ and moves to $\{2,4,6\}$ with probability $\frac{6}{8}.$ From $\{2,4,6\}:$ stay with probability $\frac{4}{8},$ move to $\{4,4,4\}$ or $\{2,2,8\}$ with probability $\frac{1}{8}$ each, and stop with probability $\frac{2}{8}.$ From $\{2,2,8\}:$ stay with probability $\frac{2}{8},$ move to $\{2,4,6\}$ with probability $\frac{2}{8},$ and stop with probability $\frac{4}{8}.$

Let $E_1, E_2, E_3$ be the expected remaining times from $\{4,4,4\},$ $\{2,4,6\},$ $\{2,2,8\}.$ Then $$\begin{aligned} E_1 &= 1 + \tfrac{1}{4}E_1 + \tfrac{3}{4}E_2, \\ E_2 &= 1 + \tfrac{1}{2}E_2 + \tfrac{1}{8}E_1 \\ &\quad {}+ \tfrac{1}{8}E_3, \\ E_3 &= 1 + \tfrac{1}{4}E_3 + \tfrac{1}{4}E_2. \end{aligned}$$ The third gives $E_3 = \frac{4}{3} + \frac{E_2}{3};$ substituting into the second yields $E_2 = 4,$ then $E_3 = \frac{8}{3}$ and $E_1 = \frac{4}{3} + E_2 = \frac{16}{3}.$

The expected number of minutes is $\frac{16}{3},$ so $m + n = 16 + 3 = 19.$