The algebraic perspective starts from constructing one $\ast$-algebra $\mathfrak{A}$ and defining states to be linear functionals $\omega:\mathfrak{A}\to\mathbb{C}$ having the property that for any $a\in \mathfrak{A}$ one has $\omega(a^\ast a)\in [0,+\infty)$ and $\omega(1)=1$. Once $\omega$ is given you can reconstruct the Hilbert space $\mathscr{H}$ and identify the appropriate inner product through a well-known construction called the GNS construction. This construction provides you also with one $\ast$-representation of $\mathfrak{A}$ on $\mathscr{H}$. The whole complexity in identifying the Hilbert space of an interacting relativistic theory lies exactly in the states $\omega$.
In more precise terms, in QFT as opposed to non-relativistic QM, one special thing happens. In non-relativistic QM all the $\ast$-representations of the $\ast$-algebra describing the canonical commutation relations are isomorphic by virtue of the Stone-von Neumann theorem. This is the well-known fact that the abstract commutation relation $[X^i,P_j]=i\delta^i_{\phantom i j}$ is implemented in a certain space of square integrable functions with one member of the pair acting by multiplication and the other by differentiation $$X^i\psi(x)=x^i\psi(x),\quad P_i\psi(x)=-i\partial_i\psi(x)\tag{1}.$$
This means that once you use this theorem the Hilbert space is in a sense identified once and for all in non-relativistic QM.
In QFT the scenario drastically changes because SvN no longer applies and you do have inequivalent representations. The states $\omega$ in $\mathfrak{A}$ are grouped together in families called folia. The follium of $\omega$, denoted $\mathfrak{F}(\omega)$ comprises all states having the property that they can be represented as density operators acting on the Hilbert space defined by the GNS construction based on $\omega$. In other words, $\chi\in \mathfrak{F}(\omega)$ whenever there is $\rho_\chi$ on the Hilbert space $\mathscr{H}_{\rm GNS}[\omega]$ such that $$\chi(a)=\operatorname{tr}(\rho_\chi\pi(a))\tag{2},$$
where $\pi$ is the $\ast$-representation of $\mathfrak{A}$ by operators in $\mathscr{H}_{\rm GNS}[\omega]$. The existence of disjoint folia in QFT means that identifying the Hilbert space is not a problem that can be decoupled from the dynamics as it happens in non-relativistic QM, and actually, such a problem is highly non-trivial. Note that if you know $\omega$ you actually know all possible vacuum expectation values of all possible operators and as a consequence you have solved the theory! In fact, for free theories we know how to identify $\omega$ and construct the associated Hilbert space. But we cannot use this Hilbert space for interacting theories.
Identifying the Hilbert space of an interacting theory is a highly non-trivial open problem in which research is actively carried out today. Personally I know of two approaches to this problem. One of them is described in this paper by Hollands and Wald and also several other papers by the authors based on the Operator Product Expansion in general quantum field theories, as appropriately axiomatized by them in the paper. The second approach is the one from the field of constructive QFT.
It relies on trying to give a construction of the Hilbert space by defining the functional measure $d\mu(\Phi)$ on a certain space of distributions using measure theory. For more details see the answer by @AbdelmalekAbdesselam in this thread and several other nice posts from him here in Phys.SE on this subject.
So the answer to:
Given a theory by its Lagrangian, exists a method to determine the corresponding inner product $\langle f, g\rangle $ inherent to that theory within the particle space encapsulated by the AQFT formalism?
Is no, there is no general prescription to identify the Hilbert space from an interacting Lagrangian. This is an actively researched subject, both in constructive QFT and in this axiomatic approach of Hollands and Wald, with possibly other approaches that I am unaware of existing as well.
