Let $S_n$ be the symmetric group over integers $[1,n]$. Let us also define an arbitrary function $f[i]$ of integer $i$.
I want to calculate $$\sum_{\sigma\in S_n}f[\sigma(i)].$$
Intuitively, I would argue that the images $\sigma(i)$ should be evenly distributed across the group; therefore, $$\sum_{k=1}^{n!}f[\sigma_k(i)]=(n-1)!\sum_{j=1}^n f[j].$$
However, that reasoning doesn't sound very formal. Is my assumption correct? How can this result be more soundly justified?
Yes, your reasoning is correct. You can first group the terms by $\sigma(i)$, which it sounds like is what you've been doing. $$\sum_{\sigma\in S_n}f[\sigma(i)]=\sum_{j=1}^nf[j]\left|\big\{\sigma\in S_n\ \colon \sigma(i)=j\big\}\right|.$$ So, all you need to show is that there are $(n-1)!$ permutations in $S_n$ for which $\sigma(i)=j$, given fixed $i$ and $j$. There are a few ways to do this -- you can explicitly count them (by repeatedly fixing values of $\sigma$). Alternatively, if you define $$T_j=\big\{\sigma\in S_n\ \colon \sigma(i)=j\big\},$$ and biject $T_j$ with $T_1$ for each $j$ (see if you can figure out such a bijection). From here, the $n$ sets $T_j$ are disjoint, of the same size, and comprise all of $S_n$, so they are each of size $|S_n|/n=(n-1)!$, as desired.