A chain in $[H,G]$ has length $\leq\Omega(|G:H|)$, where $\Omega$ denotes the number of prime factors counted with multiplicity. If $H=H_0<H_1<\dots<H_k=G$ is a maximal chain, then there are elements $g_1, \ldots, g_k$, such that $H_i=\langle H_{i-1}, g_i\rangle$. If $g_iH=g_i'H$, then $\langle g_i H_{i-1}\rangle = \langle g'_i H_{i-1}\rangle$, hence, the number of subgroup chains is bounded above by the number of sequences of cosets of $H$ in $G$, of length $\leq\Omega(|G:H|)$. Hence for $n=|G:H|$ the number of subgroups is $\leq \Omega(n) \frac{n!}{(n-\Omega(n))!} $ $ \le\Omega(n) n^{\Omega(n)}$ $\leq \log_2(n) n^{\log_2(n)} $ $ \leq n^{1+\log_2(n)}$.
To gain the missing factor 4 in the exponent you have to use the fact that there are different choices for the $g_i$, and that a group $K$ can occur in many different chains. However, this will be quite difficult. If you look at $C_p^n$, you see that most subgroups have index $\approx p^{n/2}$, thus a local argument will not work.
