This is an interesting question which I did not find answered in the literature.
Here is an example for a non-locally convex topology between the weak and the Mackey topology: The space is the Banach space $E:=L_1(0,1)$, and let $\sigma$ be the weak topology. Let $0
It remains to verify that the topology is not locally convex. For this fact I only give a sketch. The first observation is that a neighbourhood basis
of $0$ for $\tau$ is given by
$\{U_{F,\;\epsilon};\ F\subseteq(E,\|\cdot\|_1)'=L_\infty(0,1)\text{ finite},\
\epsilon>0\}$,
where
$$
U_{F,\;\epsilon}:=\{f\in E;\ \sup_{\eta\in F}|\eta(f)|<\epsilon,\
d_q(f,0)<\epsilon\}.
$$
And then one shows that the $\tau$-neighbourhood $U:=\{f\in E;\ d_q(f,0)<1\}$ of
$0$ does not contain the convex hull of any of the $0$-neighbourhoods
$U_{F,\;\epsilon}$. More precisely, one shows that the convex hull of
$U_{F,\;\epsilon}$ contains elements $f\in\bigcap_{\eta\in F}\eta^{-1}(0)$ with
$d_q(f,0)$ arbitrarily large. (This step is done in a way somewhat similar to
the proof that the topology $\tau_q$ is not locally convex.) This proves that $U$ does not contain any convex neighbourhood of $0$, and therefore the topology $\tau$ is not locally convex.