No such sequence exists. We use proof by contradiction, assuming such an infinite sequence exists, we express each $a_i$ as $a_i = 2^{r_i}b_i$, where $b_i$ is odd. Then by properties of the $\phi$-function, we have $a_i = \phi(a_{i+1}) = 2^{r_{i+1}-1}\phi(b_{i+1})$.

Note that $b_{i+1}$ cannot be $1$, otherwise we would get $b_i = 1$, $b_{i-1} = 1, \dots$, and finally $b_1 = 1$, which obviously contradicts $a_1 = (2011)!$. In this case we have $\phi(b_{i+1})$ is even. Comparing $2^{r_{i+1}-1}\phi(b_{i+1}) = 2^{r_i}b_i$ shows that $r_{i+1} \le r_i$.

Since $r_i$ are all nonnegative integers, they cannot keep decreasing forever, so there must exist some positive integer $n$ after which (below we always assume the variable $i \ge n$) we have $r_i = r_{i+1}$. Comparing the above equation shows that $4 \nmid \phi(b_{i+1})$, and
$2b_i = \phi(b_{i+1})$.

By properties of the $\phi$-function, if $\phi(b_{i+1})$ is not a multiple of 4, then $b_{i+1}$ must be a power of some odd prime, denote $b_{i+1} = p_{i+1}^{s_{i+1}}$, where $p_{i+1}$ is a prime. Now consider $b_{i+2} = p_{i+2}^{s_{i+2}}$, we have
$$b_{i+1} = \frac{\phi(b_{i+2})}{2} = \frac{p_{i+2} - 1}{2} p_{i+2}^{s_{i+2}-1} = p_{i+1}^{s_{i+1}}.$$
Hence $p_{i+2} = 3$ or $s_{i+2} = 1$.

If $p_{i+2} = 3$ (for some $i$), then $a_{i+2}$ has only the two prime factors 2 and 3, and it is easy to check that subsequently $\phi(a_{i+2}), \phi(\phi(a_{i+2})), \dots$ will also only have prime factors 2 and 3, but $a_1 = (2011)!$, a contradiction, so we must have $s_{i+2} = 1$, that is, $b_{i+2} = p_{i+2}$ (for all $i \ge n$).

Thus, for all $j \ge (n+2)$, we have $b_{j+1} = 2b_j + 1$, and these $b_j$ are all primes. We note that for any positive integer $k$, $b_{j+k} = 2^k(b_j + 1) - 1$. Let $k = b_j - 1$, then by Fermat's Little Theorem, $b_{j+k} \equiv (b_j + 1) - 1 \equiv 0 \pmod{b_j}$, hence $b_j|b_{j+k}$, which contradicts $b_{j+k}$ being prime. This completes the proof.