"Are you taller than the average of those who are taller than the average?"

We have $\mu_n \uparrow \infty$. Proof: let $$G(y) = \frac{\int_y^\infty x f(x) dx}{\int_y^\infty f(x) dx}$$ so that $\mu_{n+1} = G(\mu_n)$. Clearly $G$ is a continuous function and $G(y) > y$ for all $y$. But if $\mu_n \to \mu$ for some finite $\mu$ we must have $G(\mu) = \mu$, a contradiction.

More generally, this should show that if $X$ is a continuous random variable with essential supremum $M$, and we define $G(y) = E[X | X \ge y]$ for $y < M$, then the iterates $G^n(y) \to M$.

As in Nate's answer, we are interested in iterating the function $$G(y) := \frac{ \int_{y}^{\infty} x e^{- x^2} dx}{\int_{y}^{\infty} e^{- x^2} }.$$

The numerator is $e^{-y^2}/2$ (elementary). The denominator is $e^{-y^2}/2 \cdot y^{-1} \left( 1-(1/2) y^{-2} + O(y^{-4}) \right)$ (see Wikipedia). So $G(y) = y + (1/2) y^{-1} + O(y^{-3})$.

Set $z_n = \mu_n^2$. Then $$z_{n+1} = (\mu_n+\mu_n^{-1}/2 + O(\mu_n^{-3}))^2 = \mu_n^2 + 1 + O(\mu_{n}^{-2}) = z_n + 1 + O(z_n^{-1}).$$ So $z_n \approx n$ and we see that $\mu_n \to \infty$ like $\sqrt{n}$.

I haven't checked the details, but I think you should be able to get something like $\mu_n = n^{1/2} + O(1)$.