In QFT, fields come into play. How would the wave function of a field be? $\psi(A_\mu)$?
Somewhat analogously to the basis of position eigenstates in QM, there is a basis of field eigenstates in QFT. Each eigenstate is in correspondence to not just one value, but the configuration of the entire field, meaning all field values at all values of $x$.
I will use the real scalar field $\hat{\phi}(x)$ rather than the vector field $A_\mu$, for a simple example. The eigenstates $|\phi(x) \rangle$ are simultaneous eigenstates of every field operator $\hat{\phi}(x)$, namely for every vector $x$ we have an eigenvalue equation. Using $x=1, 3.14$ as symbolic examples (really they should be 4-vectors, so four numbers):
$$\hat{\phi}(1) | \phi(x) \rangle = \phi(1) | \phi(x) \rangle $$
$$\hat{\phi}(3.14) | \phi(x) \rangle = \phi(3.14) | \phi(x) \rangle $$
$$...$$
Here, $\hat{\phi}(x)$ denotes the operator of the scalar field. Without the hat, $\phi(x)$ is the eigenvalue corresponding to the operator at $x$ for the eigenvector $|\phi(x) \rangle $. The notation can be a bit confusing here: There is one eigenvector per field configuration, so for every possible function $\phi(x)$ we have a different eigenvector. But given a single eigenvector, it is a simultaneous eigenstate of infinitely many field operators, one for each $x$.
What role do these field eigenstates play in QFT? Not the same as $x$ in QM. You use them to derive the path integral in the theory. However, I have never seen an application in which they were used to predict predict probabilities of measurement outcomes directly through an inner product.
So, if these field eigenstates are not used to directly predict probabilities in QFT, how are probabilities predicted? Fundamentally they use a postulate from Quantum Mechanics, namely that for a system in state $|\psi \rangle$, the probability of finding $n$ particles with momentum $p_1, p_2, ... p_n$ is
$$|\langle p_1 p_2 ... p_n |\psi \rangle|^2 d^4p_1 d^4p_2 ... d^4 p_n$$
It is from this postulate that the scattering cross section is derived.
Finally, then, what about probabilities for positions? Better not ask, as there isn't a good answer to that.