Take an arbitrary route. For each face, count the number of the edges on the face which are on the route. Since they can be neither less than $3$ nor more than $4$, and they add up to $2 \times 20 = 40$, there exists exactly $4$ faces with $4$ of their edge on the route.

Now, assume that the ant passed consecutive four edges of a face as below.

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Since the ant did not pass edge $AB$, the ant passed the edges $BC$ and $AK$. Also, since all the vertices must be passed just once, looking at vertices $E$, $G$ and $I$, it follows that the route contains edges $DE$, $EF$, $FG$, $GH$, $HI$ and $IJ$. By the same reason, looking at vertices $O$ and $P$, it follows that it contains $NO$, $OP$ and $PQ$. There are two ways to complete the route: one is to take the edges $KL$, $LM$, $MN$, $CD$ and $JQ$; the other is to take $CL$, $LM$, $MQ$, $KJ$ and $DN$.

Now we can compute how many ways are there to make a route. There are $12$ ways to choose the first face, $5$ ways to choose the lacked edge, $2$ ways to complete the route, and $2$ ways to decide the direction. Since each route appears exactly $4$ times (note that there are $4$ faces of which $4$ edges are contained in a route), there are $12 \times 5 \times 2 \times 2 \div 4 = 60$ routes.