Because there are $n!$ different ways to order the hats, and $(n-1)!$ ways to distribute the hats so that persion $k$ gets his own hat, so the probability that $k$ gets his own hat is $\frac{(n-1)!}{n!}=\frac{1}{n}$
Your intuition reflects the fact that two $X_i$ and $X_j$ are not independent variables. That is a useful intuition, but it does not affect the odds of person $k$ getting his hat.
Alternatively, let's just say you are looking at $P(X_2=1)$ and person $1$ picks a hat first.
The probability that person $1$ chose person $2$'s hat is $\frac{1}{n}$. So the probability that person $2$ picks his hat amongst the rest is the probability that person $1$ did not choose that hat, times the probability that person $2$ picks the hat from the remaining $n-1$ hats. That is:
$\begin{align}P(X_2=1)= &P(\text{person 1 did not choose hat 2})\times\\ &P(\text{person 2 chooses hat 2 from the remaining }n-1\text{ hats})\\ =&\left(1-\frac{1}{n}\right)\frac{1}{n-1} = \frac{1}{n}\end{align}$