Let $u$ be the harmonic function on $\Bbb{R}^n$, we can build the following estimate using the inverse Poincare inequality:
$$\int_{B_{2r}}u^2 \ge (1+c(n)) \int_{B_r}u^2 \tag{*}$$ Where $c(n)$ is a strict positive constant.
My question is why this estimate bound the rate of decay.from the expression we see $u^2$ can not decay too fast in order to have constribution to the integral, can we deduce the explicit bound for the dacay of the harmonic function from (*)
You can not have real decay estimates, as that would imply that your function is constant. So if you would have any estimate which emplies that $u(x) \to 0$ as $|x| \to \infty$, your function would already be constant by Liousvilles theorem.
However, my best guess is to express everything in terms of $$ f(r):=\int_{B_1(0)}u(rx)^2r^n dx $$ Note that this integral is never vanishing for $r\neq 0$, since it would imply that f is equal to $0$ otherwise. Rewrite the inequality to $$ \frac{f(2r)}{f(r)}\geq 1+C \iff \ln(f(2r))-\ln(f(r))>\ln(1+C) $$ However, we can repeat that process: $$ \ln(f(2^kr))-\ln(f(r))>k\ln(1+C) $$ In particular, we can e.g. take $r=1$: $$ \ln(f(2^k))>\ln(f(1))+k\ln(1+C) $$ And taking exponential yields: $$ \int_{B_{2^k}(0)}u(x)^2dx \geq e^k e^{f(1)}e^{1+C}=C(f(1),n)e^k $$ This is expected, as e.g. multiples of your harmonic function should satisfy different growth estimates.