Solution:

Let $p_n$ be the probability that the ant is at the original vertex after $n$ minutes; then $p_0 = 1$. The chance that the ant is at each of the other three vertices after $n$ minutes is $\frac{1}{3}(1 - p_n)$. Since the ant can only walk to the original vertex from one of the three others, and at each there is a $\frac{1}{3}$ probability of doing so, we have that
$$p_{n+1} = \frac{1}{3}(1 - p_n).$$
Let $q_n = p_n - \frac{1}{4}$. Substituting this into the recurrence, we find that
$$q_{n+1} = \frac{1}{4} + \frac{1}{3}\left(-q_n - \frac{3}{4}\right) = -\frac{1}{3} q_n.$$
Since $q_0 = \frac{3}{4}$, $q_n = \frac{3}{4} \cdot \left(-\frac{1}{3}\right)^n$. In particular, this implies that
$$p_{60} = \frac{1}{4} + q_{60} = \frac{1}{4} + \frac{3}{4} \cdot \frac{1}{3^{60}} = \frac{3^{59} + 1}{4 \cdot 3^{59}}.$$