Let $n \in \mathbf N^+$. Every odd prime $\le 2n+1$ divides $2^{(2n)!} - 1$. So we have $$\varphi(2^{(2n)!} - 1) = (2^{(2n)!} - 1) \prod_{p \,\mid\, 2^{(2n)!} - 1} \left(1 - \frac{1}{p}\right) \le (2^{(2n)!} - 1) \prod_{3 \le p \le 2n+1} \left(1 - \frac{1}{p}\right)\!,$$
where $p$ is always a prime and for the first equality we have used [Euler's product formula][1]. It follows that the limit inferior in the OP is $0$.


  [1]: https://en.wikipedia.org/wiki/Euler%27s_totient_function#Euler.27s_product_formula