So let us start with the "simplest" scheme over $Spec(\mathbf{Z})$ namely $X\_0=Spec(\mathbf{Z})$. Then the (reciprocal) Weil zeta function of $X\_0$ at a prime $p$ is given by $Z\_p(T)=1-T$ (a polynomial of degree $1$). So the Hasse-Weil zeta function of $X\_0$ is given by
$$
L\_{X\_0}(s):=\prod\_{p}Z\_p(p^{-s})^{-1}=\zeta(s).
$$
Now if one lets $\psi(z)=\sum\_{n\in\mathbf{Z}}e^{i\pi n^2z}$ then $\psi(z)$ is a modular form of weight $1/2$ (over a suitable congruence group of $SL\_2(\mathbf{Z})$) in the sense that
$$
(-iz)^{-1/2}\psi(-1/z)=\psi(z). \;\;\;\;(\*)
$$
A straight forward computation which uses the definition of the Gamma function implies that
$$
\tilde{L}\_{X\_0}(s):=\int\_{0}^{\infty}(\psi(it)-1)t^{s/2}\frac{dt}{t}=2\pi^{-s/2}\Gamma(s/2)\zeta(s).
$$
Using the functional equation $(\*)$ one may deduce the meromorphic continuation and the functional equation of $\zeta(s)$ (invariance of $\tilde{L}\_{X\_0}(s)$ under $s\mapsto 1-s$).


Now let us take the scheme $X\_1=\mathbf{P}^1$ over $Spec(\mathbf{Z})$. Then the (reciprocal) Weil zeta function of $X\_1$ at a prime $p$ is given by $Z\_p(T)=(1-T)(1-pT)=1-\sigma\_1(p)T+pT^2$ (a polynomial of degree $2$). It thus follows that the 
Hasse-Weil zeta function of $X\_1$ is given by
$$
L\_{X\_1}(s)=\prod\_{p}Z\_p(p^{-s})^{-1}=\zeta(s)\zeta(s-1).
$$
Now let us look at the Eisenstein series of weight $2$ i.e.
$$
E\_2(z)=(2\pi i)^{-2}\sum\_{m,n}'\frac{1}{(mz+n)^2}:=\frac{-B\_2}{2}+2\sum\_{n\geq 1}\sigma\_1(n)q\_{z}^n,
$$
where $q\_{z}=e^{2\pi iz}$. (Note that I don't get any convergence issue here since I take this $q$-expansion as the definition of $E\_2(z)$). Note that $E\_2(z)$ is "almost" a moldular form of weight $2$ (for the full congruence group $SL\_2(\mathbf{Z})$) since 
$$
(-z)^{-2}E\_2(-1/z)=E\_2(z)-\frac{1}{2\pi iz} \;\;\;\; (\*\*)
$$
A straight forward computation similar to the one before implies that
$$
\tilde{L}\_{X\_1}(s):=\int\_{0}^{\infty} (E\_2(it)+B\_2/2)t^{s}\frac{dt}{t}=2 (2\pi)^{-s}\Gamma(s)\zeta(s)\zeta(s-1).
$$
As before using $(\*\*)$ one obtains the meromorphic continuation of $L\_{X\_1}(s)$ and its functional equation ($\tilde{L}\_{X\_1}(s)=-\tilde{L}\_{X\_1}(2-s)$, note the appearance of the sign $-1$). Note that this could already be deduced from what we know from $L\_{X\_0}(s)$.


Now there is no reason to stop here. So let $X\_2=\mathbf{P}^2$ over $Spec(\mathbf{Z})$.
Then the Weil zeta function of $X\_2$ at $p$ is $Z\_p(T)=(1-T)(1-pT)(1-p^2T)$ (a polynomial of degree $3$). It thus follows that the 
Hasse-Weil zeta function of $X\_2$ is given by
$$
L\_{X\_2}(s)=\prod\_{p}Z\_p(p^{-s})^{-1}=\zeta(s)\zeta(s-1)\zeta(s-2)=\sum\_{n\geq 1}\frac{a\_n}{n^s}
$$


Q1: Is it reasonable to expect the formal $q$-expansion $f(q\_z)=\sum\_{n\geq 1} a\_n q\_z^n$ to be related in some direct way to an automorphic form w.r.t. a suitable congruence subgroup of $GL\_3(\mathbf{Z})$?


Q2: What about $X\_n=\mathbf{P}^n$ in general?


**added**: Note that in the case of $X\_0$ I'm really looking at $\tilde{\zeta}(s):=\zeta(2s)$ which is an $L$-function of weight $1/2$ in the sense that $\tilde{\zeta}(s)$ is related to $\tilde{\zeta}(1/2-s)$ which is in accordance with the fact that $\psi(z)$ has weight $1/2$.



On the Hasse-Weil L-function of $P^n$
$\newcommand\GL{\mathrm{GL}}$
$\newcommand\SL{\mathrm{SL}}$
$\newcommand\R{\mathbf{R}}$


There is, of course, an autormorphic representation (Hecke character)
$\chi$ for $\GL(1)$ whose $p$-adic avatar is the cyclotomic character. From this, one thus has the isobaric sum (following Langlands and Jacquet-Shalika) 
$$\pi = 1 \boxplus \chi \boxplus \ldots \boxplus \chi^{n-1}$$
which is an automorphic representation for $\GL(n)$ with $L(\pi,s) = L(\mathbf{P}^n,s)$. 


But all of this is somewhat irrelevant to your action question, to which the answer is *not really*. The reason $q$-expansions arise for $\GL(2)$ has to do with the fact that
$\SL\_2(\R)/\mathrm{SO}\_2(\R)$ is the upper half plane, and $\SL\_2(\mathbf{Z})$ contains the
element $z \mapsto z+1$ for which $q = e^{2 \pi i z}$ is invariant. 
$\SL\_n(\R)/\mathrm{SO}\_n(\R)$ is quite a different beast.


 **Edit** You don't seem satisfied with my answer, but I think you seem to be missing a basic principle:  **if you write down something at random, there's no reason it should be interesting.**  The theory of automorphic forms, however labyrinthian, has an incredibly precise structure. If you take an automorphic representation $\pi$ corresponding to a classical modular form, then the representation theory of $\mathrm{GL}\_2(\mathbf{R})$ will
tell you that a lowest weight vector for the discrete series will be annihilated by some differential operator which one can compute to be the Cauchy-Riemann equations, and hence
the corresponding function on the upper half plane will be holomorphic: this is a very
specific reason why holomorphic functions might be associated to automorphic forms. If, instead, one works with an automorphic form for $\mathrm{GL}(3)$, then the representation theory will (under appropriate conditions) produce an expansion in terms of Whittaker functions which will have a completely different flavour. 


It might also be worth remarking that already since Langlands time people have thought hard about the problem of  *transfer*, that is, for example, starting with an automorphic representation for $\GL(2)$ and producing one for $\GL(3)$. The methods used to prove these results are almost exclusively via the trace formula - in particular, they proceed via harmonic analysis and representation theory, rather than explicit manipulations with functions (holomorphic or otherwise). Thus (addressing your comment) the hope that one
might explicitly decompose the symmetric space of $\GL(3)$ in some way is a little too optimistic. 


Finally, you can (of course) recover $L(\pi,s)$ from $f(q)$, so $f(q)$ does carry (in some sense) all the information of $L(\pi,s)$. Moreover, one can recover $f(q)$ as the inverse Mellin transform of
$L(\pi,s)$ adjusted by appropriate Gamma factors. However, you will find that when $n > 2$ the relevant Gamma factors involved will not play well with respect to the functional equation, and thus there will be no reason to expect that $f(q)$ will not have any nice properties (such as a nice functional equation).
Even for $n = 1$ you have to cheat a little (replacing $q^n$ by $q^{n^2}$),
and moreover the theta function is (automorphically) more naturally thought of
as associated to
the metaplectic group rather than $\GL(2)$.


