Let $n \in \mathbf N^+$ and $a \in \mathbf N_{\ge 2}$. Every prime $\le 2n+1$ that doesn't divide $a$, is a divisor of $a^{(2n)!} - 1$ (by Fermat's little theorem). So we have $$\frac{\varphi(a^{(2n)!} - 1)}{a^{(2n)!} - 1} = \prod_{p \,\mid\, a^{(2n)!} - 1} \left(1 - \frac{1}{p}\right) \le \prod_{a < p \le 2n+1} \left(1 - \frac{1}{p}\right)\! \stackrel{n \to \infty}{\longrightarrow} 0,$$ where $p$ is always a prime and for the first equality we have used Euler's product formula. In particular, this shows that the limit inferior in the OP is $0$.

Let $n \in \mathbf N^+$ and $a \in \mathbf N_{\ge 2}$. Every prime $\le n+1$ that doesn't divide $a$, is a divisor of $a^{n!} - 1$ (by Fermat's little theorem). So we have $$\frac{\varphi(a^{n!} - 1)}{a^{n!} - 1} = \prod_{p \,\mid\, a^{n!} - 1} \left(1 - \frac{1}{p}\right) \le \prod_{a < p \le n+1} \left(1 - \frac{1}{p}\right)\! \stackrel{n \to \infty}{\longrightarrow} 0,$$ where $p$ is always a prime and for the first equality we have used Euler's product formula. In particular, this shows that the limit inferior in the OP is $0$.

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 $0$.

Let $n \in \mathbf N^+$ and $a \in \mathbf N_{\ge 2}$. Every prime $\le 2n+1$ that doesn't divide $a$, is a divisor of $a^{(2n)!} - 1$ (by Fermat's little theorem). So we have $$\frac{\varphi(a^{(2n)!} - 1)}{a^{(2n)!} - 1} = \prod_{p \,\mid\, a^{(2n)!} - 1} \left(1 - \frac{1}{p}\right) \le \prod_{a < p \le 2n+1} \left(1 - \frac{1}{p}\right)\! \stackrel{n \to \infty}{\longrightarrow} 0,$$ where $p$ is always a prime and for the first equality we have used Euler's product formula. In particular, this shows that the limit inferior in the OP is $0$.