Collecting my comments into an answer:

The rationals are countable, so all the sets $L_n$ are countable, so their union is countable, but the complex numbers are uncountable, so not every complex number is in at least one of the sets. That being said, I'm not sure there's even a single number $x$ of which one could say that it has been proven that $x$ is not in the union of the $L_n$.

For a number $x$ to be in more than one of the sets, there would have to be a rational number $y$ such that at least one of the numbers $\exp(y),\exp(\exp(y)),\exp(\exp(\exp(y))),\dots$ is rational. It is known that $y$ and $e^y$ are both rational if and only if $y=0$. To the best of my knowledge, no one knows whether there is any rational $y$ such that some iterated exponential of $y$ is rational. For example, I believe it is unknown whether $e^e=\exp(\exp(1))$ is rational.