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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. It follows that the limit inferior in the OP is $0$.