$$f(x_1) = n,\ f(x_2) = n^2,\ f(x_3) = n^3.$$
Then for a polynomial $g(x) = f(x) - n$ we have $g(x_1) = 0$, $g(x_2) = n^2 - n$, $g(x_3) = n^3 - n$, so $g(x) = (x - x_1)(x - t)$ for some integer $t$. Then we need to find $x_1, x_2, x_3, t$, such that $g(x_2) = (x_2 - x_1)(x_2 - t) = n^2 - n$, $g(x_3) = (x_3 - x_1)(x_3 - t) = n^3 - n$. It suffices to have
$$x_2 - x_1 = 1,\ x_2 - t = n^2 - n,\ x_3 - x_1 = n,\ x_3 - t = (n^2 - 1).$$
For example, $x_1 = 0$, $x_2 = 1$, $x_3 = n$ and $t = -n^2 + n + 1$ satisfy these relations. It's easy to check that the polynomial $f(x) = x(x + n^2 - n - 1) + n$ attains the desired values at $x = 0$, $x = 1$, and $x = n$.