You can just compute the spectrum exactly using the formula mentioned by Benjamin Steinberg: $$\sum_g \operatorname{ord}(g) \frac{\chi(g)}{\chi(1)}.$$ There are two or four characters of degree $1$ according to the parity of $n$, and these obviously give an integer value since $\chi(g) = \pm1$ and $\chi(1) = 1$. The characters of degree $2$ have the form $r^i \mapsto \zeta^i + \zeta^{-i}$ where $\langle r\rangle$ is the rotation subgroup and $\zeta \ne \pm1$ is an $n$th root of unity. The value of the character on $G \setminus \langle r \rangle$ is zero. Therefore the sum above is $$\sum_{i=0}^{n-1} \frac{n}{\gcd(i, n)} \frac{\zeta^i + \zeta^{-i}}2 = \sum_{i=0}^{n-1} \frac{n}{\gcd(i, n)} \zeta^i.$$ This sum is invariant under the action of the Galois group and therefore a rational algebraic integer, a.k.a., an integer. You can compute its value exactly if you want using the fact that the sum of the primitive $d$th roots of unity is $\mu(d)$.
The observation that the sum is an integer for other finite groups sounds more interesting. I wonder if it is related to the result of Frobenius that the number of solutions to $g^n = 1$ is a multiple of $n$.
Your second question is about how $n = n(p, q)$ depends on $p$ and $q$. This is completely unrelated and a bit unnatural. One may be able to model $f_p f_q$ as the product of random involutions supported on $\{1, \dots, p-1\}$ and $\{1, \dots, q-1\}$.
