You are of course right that the argument is not complete as it stands. It may be possible to fix it by observing that $H^i_{\mathrm{un}}$, being a subspace of $H^i$, is effaceable; so if it were a $\delta$-functor it would be the universal one.

Another, cheaper way of seeing this: $H^i_{\mathrm{un}}$ and $H^i$ agree in degree 0, so if they both satisfied a long exact sequence, they'd have to agree in all degrees which is not true.

I’m afraid the question is now too vague to be meaningfully answerable.

Moreover, if X(N) is that quotient viewed as a Q-var, the covering group of X(p q) over X(q) for primes p, q is GL2(Fp), not SL2(Fp) as you seem to want.

Or are you happy for the C-points of C(Gamma) to be a disjoint union of multiple copies of Gamma \ H?

The canonical model of that quotient is a non-geometrically-connected curve over Q whose connected components are defined over Q(zeta_N). So it does not give you a variety over Q whose C-points are Gamma \ H, contrary to your claim.

For this version of the BB conjecture, it doesn't matter whether you include the Gamma factors at infinite primes or not, since they cannot vanish at s = i. (This is easy to see from the fact that $H^{2i-1}(X)$ is a pure motive of odd weight, cf the recipe for the Gamma factors in Deligne's "Valeurs de fonctions L et periodes d'integrales".)

The theory of Shimura varieties doesn't do what you claim it does, I'm afraid. If $\Gamma$ is a congruence subgroup of $SL(2, \mathbb{Z})$ of level $N$, then you always get a canonical model of $\Gamma \backslash \mathcal{H}$ over $\mathbb{Q}(\zeta_N)$, but it will not necessarily descend all the way to $\mathbb{Q}$. Even if it does descend, it will not descend "canonically", so you cannot expect this to be compatible with the action of $\Gamma / \Gamma'$ on $C(\Gamma')$ for $\Gamma' \trianglelefteq \Gamma$.

No, I do not know the result if $a_i \to R$ (absent any control on the rate of convergence).

This isn't clear to me either. I suggest you ask the authors.

It will work for $n = 4$ but the statement will be a little more fiddly. With cubes, either $a$ is a cube mod p or it isn't; but for squares there are 3 possibilities, $a$ can be a 4th power, a square but not a 4th power, or a non-square mod p. Try starting from the sentence "A weak form of Cubic Reciprocity is ..." at the bottom of p160 of Diamond & Shurman and working out the quartic analogue of that statement.

Try thinking about this in terms of Galois representations. Where you have 2-dimensional representations of Galois groups it is reasonable to look for modular forms. If $N > 4$ then you will be seeing higher-dimensional representations, which might have a parametrisation by some more complicated automorphic object, but will not come from modular forms.

This question is far too broad; voting to close.