Consider the involution of $S_n$ which sends a permutation $\sigma$ to the permutation $\sigma^*$ defined by $\sigma^*(i) = j$ if and only if $\sigma(n-j+1) = n-i+1$. Notice that, for each $\sigma$ in $S_n$, the assignment $i \mapsto n-\sigma(i)+1$ defines a bijection from $I(\sigma)$ to $I(\sigma^*)$ whose inverse sends $j$ to $n-\sigma^*(j)+1$. Consequently,
$$\begin{align*} \sum_{\sigma \in S_n} \frac{1}{|I(\sigma)|} \sum_{i \in I(\sigma)} (i + \sigma(i)) &= \\ &= \frac{1}{2} \sum_{\sigma \in S_n} \left( \frac{1}{|I(\sigma)|} \sum_{i \in I(\sigma)} (i + \sigma(i)) + \frac{1}{|I(\sigma^*)|} \sum_{i \in I(\sigma^*)} (i + \sigma^*(i)) \right) \\ &= \frac{1}{2} \sum_{\sigma \in S_n} \frac{1}{|I(\sigma)|} \sum_{i \in I(\sigma)} (i + \sigma(i) + (n - \sigma(i) + 1) + (n - i + 1)) \\ &= (n + 1)! . \end{align*}$$