I would like to compute $\mathbb{E}[X_{1} \mid X_{1} + \cdots X_{n} \ = x]$ if $X_{1}, \ldots, X_{n}$ are i.i.d random variables.
Intuitively, I think that the answer should be $x/n$ because there are $n$ random variables with the same distribution. If they each contribute the same amount, then we would have $x/n \cdot n = x$, which is what we want the sum to equal.
How can I show this, though?
Your idea is exactly right. To express it rigorously, let $S_n=X_1+...+X_n$. The key observation is the following
Claim: $E(X_1|S_n)=E(X_i|S_n)$ for any $i$.
To see this, let $\sigma$ be any permutation of the indices $1,\dots, n$. Let $\tilde{X_i}=X_{\sigma(i)}$ and let $\tilde{S_n}=\sum_i \tilde{X}_i$. Note that the random vector $(X_1,\dots, X_n)$ has the same distribution as the random vector $(\tilde{X_1}, \dots, \tilde{X_n})$ (by iid-ness). By applying the measurable function $(x_1, \dots,x_n)\to (x_1, \sum_ix_i)$ to each of these random vectors,we conclude that $(X_1,S_n)=^d(\tilde{X}_1,\tilde{S_n})$. In particular, $E(X_1|S_n)=E(\tilde{X}_1|\tilde{S_n})$. On the other hand, $S_n=\tilde{S_n}$ and $\tilde{X}_1=X_{\sigma(1)}$. So $E(X_1|S_n)=E(X_{\sigma(1)}|S_n)$. But $\sigma$ is arbitrary, so $\sigma(1)$ could be any index $i$. This establishes the claim.
The rest of your argument goes through essentially unchanged. Noting that$\sum_i E(X_i|S_n)=E(S_n|S_n)=S_n$,and applying the Claim, we get $nE(X_1|S_n)=S_n$ as desired.