Prof. Will Sawin, Thanks a lot. I still can not understand you last paragraph. Could you explain or sketch the the Euler factor as you said ".....you'll get at each prime $q$ other than $P$ an Euler factor that is..."? This is really what I need. If promised, could I send an E-mail talking this issue? Much obliged!

So if there is a way to give the $L$-function in terms of the Fourier coefficients?

I wander what is the exact factors between $L(s,f\times g \times h)$ and the Dirichlet series $\sum_{n\ge 1} \frac{\lambda_f(n)\lambda_g(n)\lambda_h(n)}{n^s}.$ The Gamma factor is known to be $L_\infty(s,f\times g \times h)=(2\pi)^{-2s}\Gamma( s)\Gamma (s+1-k)\Gamma ( s+1-k^\prime)\Gamma ( s+1-\kappa)$, where $k,k^\prime, \kappa\ge 2$ are the weights of respective newforms.

Dear prof. Will Sawin, thanks for your timely reply. well, I cannot catch you a bit. Could you point the exact factor between $\zeta^{(MN)}(2s)$ and $\sum_{n\ge 1}\frac{\lambda_f(n)\lambda_g(n)\lambda_h(n}{n^s}$ ?

My concern is that how large of the level $N$ compared to the length $X$, such that we have a lower bound like $\sum_{p\le x}|\lambda(p)|^2\gg x/\log x$. Note that one needs the $\gg$-constant no longer depends on the level. However, it seems there is no a definite answer towards this in the literature.

So it is urgent to figure out the minimum value of $c_1$ to ensure that $c_2 \frac{x}{\log x} \le \sum_{p\le x}|\lambda(p)|^2 \le c_3 \frac{x}{\log x}$.

So for any given $c_1$, whether or not one could has $\displaystyle c_2 \frac{x}{\log x} \leq \sum_{p\leq x}|\lambda(p)|^2\leq c_3 \frac{x}{\log x},$ where $c_2,c_3$ are computable depending on $c_1$?

@Will Sawin Dear Prof. Will Swain, thanks for the heuristic guide. I will try to pursue this vein.

Many thanks! A naive question is whether or not the constants $c_2,c_3$ respectively depend on $c_1$? Following P. Humphries-J. Thorner's paper, it looks like the constant $c_6$ in the final estimate in Theorem 2.1 depends on the former given constants $c_1,c_2,c_3,c_4,c_5$? By the way, the estimate $c_2 \frac{x}{\log x}\le \sum_{p\le x} |\lambda(p)|^2\le c_3\frac{x}{\log x} $ definitely does not depend the form $f$ anymore.

@Will Sawin Dear Prof. Will Swain, thanks for comments. Much obliged!