What form of explicit description do you have in mind?
If you let $n$ vary, any finitely generated field of extension of $k$ can occur as an $A(\mathfrak p)$,
as is easily seen, and so classifying the possible $A(\mathfrak p)$ is the same as classifying the possible f.g. field extensions of $k$, which is essentially the same thing as classifying algebraic varieties over $k$ up to birational equivalence, a fairly difficult problem.
One thing you can say is that the transcendence degree of $A(\mathfrak p)$ over $k$ is the equal to the Krull dimension of $A/\mathfrak p$, which in turn is equal to $n - \mathrm{height} \, \mathfrak p$.
Let's consider the case $n = 2$, as an example. Then $\mathfrak p$ is either maximal
(equivalently, $A(\mathfrak p) = k$), the zero ideal (equivalently,
$A(\mathfrak p) = k(x_1,x_2)$), or height one (equivalently, $\mathfrak p$ is principal). In the height one case, the field $A(\mathfrak p)$ is the function field of an algebraic curve over $k$.
Any f.g. field extension of $k$ of transcendence dim'n $1$ can arise as
an $A(\mathfrak p)$, and for a fixed such extension $K$, the classification of the
possible $\mathfrak p$ for which $A(\mathfrak p) \cong K$ is equivalent
to the classification of the (possibly singular) plane models of the curve $C$ corresponding to $K$.
The classifiation of the different possible $K$ is the problem of classifying all curves over $k$, which is an elaborate theory (involving the concept of genus of a curve, the moduli space of curves of a given genus, and so on).
Just to be even more specific, if $\mathfrak p = (f)$ with $f$ of degree one or two, then $A(\mathfrak p) \cong k(x).$ But in fact there are irreducible $f$ of every degree for which $A\bigl((f)\bigr) \cong k(x)$, although for most $f$ of degree $> 2$, the field $A\bigl( (f)\bigr)$
will not be isomorphic to $k(x)$. (E.g. if $f$ is of degree three, then the condition for $A\bigl((f)\bigr)$ to be isomorphic to $k(x)$ is that the homogenization of $f$ describe a singular cubic.)
