The phrase "essentially square-integrable" confused me. I thought it is the same as "tempered". Actually, it means "square-integrable mod center". So my previous comment is not correct.

Is it because of square-integrability vs. $\textit{essentially}$ square-integrability? An essentially square-integrable representation can be unramified.

Indeed, the center of the smaller group is the only problem. So one may hope that a ``naive'' regularization, that is, regularizing only the central direction would work. This is what we have done in our recent preprint, \S4.6 arxiv.org/abs/2111.02297v1.

One way to think about the Fourier--Whittaker expansion is by adelizing the automorphic form so that there is ``only one cusp'' to think about. Given the congruence subgroup one may adelize the space via strong approximation and the automorphic form to consider forms on $\mathrm{GL}_n(\mathbb{A})$ invariant under some hyperspecial open compact subgroup $K_f(N)$. Then understanding the local Whittaker functions is enough to obtain the full Fourier expansion adelically which one may re-understand in classical language.

Typically, no. Because, square of an Eisenstein series will have non-zero components in most representations in the spectral decomposition. Take a cusp form $\phi$ and a maximal parabolic Eisenstein series $E$ on $\mathrm{GL}(n)$. If there were an Eisenstein series $\mathbb{E}=E^2$ then $$0 = \langle \phi,\mathbb{E}\rangle = \langle \phi\bar{E},E\rangle.$$ However, the last inner product is the global zeta integral of $\phi\otimes E$ and typically does not vanish identically.

What do you mean by ''$f$ has small support near $e$''? By definition, $f(n)=f(e)$, for $n\in N$, which means that $f$ can not have compact support in $G$. If you meant to say ''small support'' in $N\backslash G$ that also seems unlikely. Because $f(nm)=\sigma(m)v$ for $m\in M$. This is nonzero (for $v\neq 0$), for instance, if $M$ is a torus.

Sorry I am not much familiar with this $M_{1,w}$ operator. Does it come from the Laurant expansion around $s=0$? How does the coefficient of the expansion depend on $s$ then?

Intertwining operators are not holomorphic. They have (typically) simple poles at the ``points of reducibility''. For instance, for $\mathrm{SL}(2)$ they appear at integer $s$ and give rise to special representations (by sub or quotient representation). $M(s,\sigma) M(-s,w\sigma)=Id$ should be interpreted as if $M(s,\sigma)$ has a pole then $M(-s,w\sigma)$ has a zero, i.e. has a non-trivial kernel.

Is $\phi$ $K$-finite on the left? Or right?

If you are asking for Ramanujan on average uniformly in $n$, to prove that it will be as hard as proving Ramanujan: Let $n$ be any large prime bigger than $x$ so that $(m,n)=1$ for all $m<x$ so that $$\lambda(n,m)=\lambda(n,1)\lambda(1.m).$$ If we have $$\sum_{m<x} |\lambda(n,m)|^2\ll x^{1+\epsilon}$$ uniformly in $n$ then using the fact that $$\sum_{m<x}|\lambda(1,m)|^2\gg x^{1-\epsilon}$$ we obtain $$\lambda(n,1)\ll x^\epsilon < n^\epsilon.,$$ i.e. Ramanujan.

I thought that $n$ in the second sum is fixed, say $n_0$. Then the second sum is bounded by $\sum_{n^2m<n_0^2x} |\lambda(n,m)|^2\ll_{n_0} x^{1+\epsilon}$ from the first estimate in your question.

If $\alpha_{p,j}:=p^{\mu_{p,j}}$ are bounded independent of $p$ then $\Re(\mu_{p,j})\le 0$ for all $j$. At least for $PGL(n)$ this implies that $\Re(\mu_{p,j})=0$ for all $j,p$, i.e. Ramanujan.

I was originally thinking about the analytic conductor, but if you like you may include either of them.