Forgive my limited capaicty if the mistakes are making. I think $dominate$ maybe depends on the choices of $a_m$. Note the integral with respect to $t$ is $\le |\log^{-1}(n/m)|$. For off-diagonal terms, $m,n$ can be restricted to $n>m,|m-n|=\Delta$ with $\Delta\le m\le M$. The off-diagonal terms can be bounded by $\ll \sum_{m\le M}m |a_m| \sum_{\Delta\le m}|a_{m+\Delta}|/\Delta$ which is big if $a_m>0$ for any $m$, and one of $a_m$ for $M<m\le 2M$ is sufficiently large, $T/\epsilon^2$, say.

I search on Google that (the paper matwbn.icm.edu.pl/ksiazki/aa/aa84/aa8426.pdf by Goldston and Gonek) in the truncated interval $[\alpha T,\beta T]$ the asymptotic (Theorem 1) was obtained under the restrictions (6),(7),(8). This shows that the low bound is closely related to the propeties of the real sequences $a_n.$ I also feel that when $M\le\epsilon T $, "the diagonal terms dominate" may be inaccuate.

@Lucia Thanks. I was informed by Brad Rodgers the copy can be downloaded in a website (lib.freescienceengineering.org/view.php?id=807106). Much obliged for timely reply, good day.

$(3.3)$ in Andrew Granville paper in my comment implies when $q$ is small(perhaps $x>q^{1+\epsilon}$ can satisfies) $$\pi(x,q;a)\sim \frac{x-x^\beta}{\varphi(q)\log x}\ge (1-\epsilon) \frac{x}{\varphi(q)\log x}$$ where $\beta$ is real zero of $L(s,\chi) $ that is close to $1 $ with the assumption $\chi(a)=1 $ which can be omitted.

Brilliant! Thanks for your reply. In IK'book, $(5.20)$ states $L(\sigma+it,f)\ll_{\epsilon,f}C(\frac{1}{2}+it,f)^{\frac{1-\sigma}{2}+\epsilon}$, when $0\le\sigma\le 1$ and $t$ is large. $C^\epsilon$ factor may be not omitted in the convexity bound above.