Question: i know that|$S\_n$|=$n!$ and |$A\_n$|= $\frac{n!}{2}$ but i'm not sure where to go from there. Thanks.

Answer: 
You might find it easier to label the moves $N,E,S,W$, so a path might look like $NNESWS$. You're effectively counting the number of six-letter strings where the number of $N$s is the same as the number of $S$s, and same for $E/W$.

To do this systematically: what if the movement is only North/South? In other words, how many ways can we arrange the letters $NNNSSS$?

The next case would be four North/South moves and two East/West: $NNSSEW$.

The next cases will have the same counts as above, by symmetry. (Swap all $N/S$ with $E/W$.)
This is a minor refinement on @Theophile's answer.

Any closed path must have the same number of $N$s and $S$s, and the same number of $E$s and $W$s. So we can consider them in unordered pairs $(NS)$ and $(EW)$. Any closed path must have just three pairs. Call $k$ the number of $(NS)$ pairs.
Thus $0\leq k \leq 3$. For each of these cases the number of $(EW)$ pairs is determined and is $3-k$.

For each $k$, you can determine the number of steps having the specified $k$ $N$s by ${6 \choose k}$ for the $N$, and the specified $k$ $S$s by ${6-k \choose k}$, and ${6 - 2k \choose (6-2k)}$ for the $E$, and ${6 - 2k - (2k) \choose (6-2k)}$ for the $W$. (Of course only non-negative terms should be included. Multiply these for a given $k$.

Now sum over all possible values of $k$.