$\newcommand\GL{\mathrm{GL}}$ $\newcommand\SL{\mathrm{SL}}$ $\newcommand\R{\mathbf{R}}$

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.

Finally, you 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. (Even for $n = 1$ you have to cheat a little, the naive thing one would write down would be $\sum q^n = q/(1-q)$, which is a nice function but doesn't have much to do with automorphic forms.)

$\newcommand\GL{\mathrm{GL}}$ $\newcommand\SL{\mathrm{SL}}$ $\newcommand\R{\mathbf{R}}$

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 $\GL(2)$.