No, not efficiently, not in general. Suppose $p=2p'+1$ and $q=2q'+1$ where $p',q'$ are prime. Then factoring $pq$ is believed to be hard. (Indeed, these primes are known as safe primes, and factoring a product of two safe primes is believed to be hard.) However we have
If you could compute $\varphi(\varphi(pq))$ from $pq$ efficiently for $p,q$ of this form, then you could factor $pq$ efficiently for $p,q$ of this form. The reduction works as follows. Consider the quadratic function $f$ given by
$$f(x)=(x-p')(x-q')=x^2 -(p'+q') + p'q'.$$
We can compute the coefficients of $f$, as
so you can use the quadratic formula to solve for the roots of $f$ and recover $p',q'$. From this the factorization of $pq$ can be recovered.
So, no, you are not overlooking anything. There is likely no efficient way to compute $\varphi(\varphi(pq))$ for arbitrary numbers $pq$ (unless factoring is easy).