For $n\geq 5$, all the eigenvalues of the Cayley graph you describe are integers. The eigenvalues correspond to character values on $3$-cycles multiplied by the number of $3$-cycles (because the generating set of $3$-cycles is a conjugacy class). The multiplicity is the dimension of the irreducible representation.

Since the character values are algebraic integers, it is enough to observe that on a $3$-cycle all character values are rational. This follows from the well-known fact (see Noam Elkies answer to this question) that if $g$ is an element of a finite group $G$, then $\chi(g)$ is rational for all characters $\chi$ of $G$ if and and only if $g^m$ is conjugate to $g$ for all powers $m$ coprime to the order of $g$.

Since every power of a $3$-cycle that is not the identity is a $3$-cycle and $3$-cycles are all conjugate in $A_n$ for $n\geq 5$, this gives the integrality of character values on $3$-cycles.

For $n\geq 5$, all the eigenvalues of the Cayley graph you describe are integers. The eigenvalues correspond to character values on $3$-cycles multiplied by the number of $3$-cycles (because the generating set of $3$-cycles is a conjugacy class). The multiplicity is the dimension of the irreducible representation.

Since the character values are algebraic integers, it is enough to observe that on a $3$-cycle all character values are rational. This follows from the well-known fact (see Noam Elkies answer to this question) that if $g$ is an element of a finite group $G$, then $\chi(g)$ is rational for all characters $\chi$ of $G$ if and and only if $g^m$ is conjugate to $g$ for all powers $m$ coprime to the order of $g$.

Since every power of a $3$-cycle that is not the identity is a $3$-cycle and $3$-cycles are all conjugate in $A_n$ for $n\geq 5$, this gives the integrality of character values on $3$-cycles.