The quantity $\phi(n)$ grows like $\Theta(n)$.
For the lower bound, consider the following graph: a clique on $m$ vertices connected to a path of length $k = C\log m$ for an appropriate $C$ (so $n = m + k \approx m$), in which each vertex of the path is also connected to the first vertex of the path (the one incident to the clique). The maximal hitting time is some large number $M = \exp(k)$. The hitting time of two vertices in the clique is around $m \ll M$. The hitting time of a vertex on the path and any other vertex is at most $M$. So the average hitting time is at most roughly $m + \frac{2kn}{n^2} M = O(kM/n)$. In fact, the hitting time of a vertex at distance $t$ from the clique is roughly $\exp(t)$ (for $t \gg \log m$), and if we take this into account then we get an average hitting time of roughly $m + \sum_{t=1}^k \frac{2n}{n^2} \exp(t) = O(M/n)$. This shows that $\phi(n) = \Omega(n)$.
For the upper bound, let $H_{ij}$ be the maximum hitting time. The triangle inequality shows that for each $k$, either $H_{ik} \geq H_{ij}/2$ or $H_{kj} \geq H_{ij}/2$. So at least $n/2$ of the hitting times are at least $H_{ij}$, showing that the average hitting time is at least $H_{ij}/n$. This shows that $\phi(n) = O(n)$.
