The $q$-binomial ${n+k \choose k}_q$ is a polynomial in $q$. The coefficient of each power $t \leq nk$ of $q$ gives the number of lattice paths from $(0,0)$ to $(n,k)$ which enclose an area $t$ beneath.
The $q$-multinomial ${n_1+n_2+\dots+n_N \choose n_1,n_2,\dots,n_N}_q$ is also polynomial in $q$. When $q=1$, as expected, you get the number of lattice paths from the origin of the hypercuboid of dimension $n_1 \times \dots \times n_N$ to the vector $(n_1,\dots,n_N)$. For example: $${8 \choose 3,3,2}_1=560$$ But! Notice this... $${8 \choose 3,3,2}_q = 1 + 2 q + 5 q^2 + 9 q^3 + 15 q^4 + 22 q^5 + 31 q^6 + 39 q^7 + 47 q^8 + 53 q^9 + 56 q^{10} + 56 q^{11} + 53 q^{12} + 47 q^{13} + 39 q^{14} + 31 q^{15} + 22 q^{16} + 15 q^{17} + 9 q^{18} + 5 q^{19} + 2 q^{20} + q^{21}$$
Why is there a $q^{21}$ term? Surely, if the "area under a multidimensional lattice path" has any meaning, the maximum area you can enclose (only one lattice path can achieve this) is $3 \times 3 \times 2 = 18$.
To make this area (i.e. volume) enclosed in higher dimensions clearer, this is an example case for ${6 \choose 2,2,2}_q$, where the area enclosed is 2.
Here, it is 4:
Is there a generating function for the area under a lattice path in $d$-dimensions (i.e. on a hypercubic lattice)? The lattice path enumeration books I see all stop at that point. They can count the paths in higher dimensions, but seem to leave out higher dimensional analogues of the $q$-analogue representation of the hypervolume underneath.
It is equivalent to partitioning an integer into $k$ parts, each less than $m$, but with I think restricted parts. For example, their are certain areas you cannot enclose with a lattice path in some cuboids, hence the polynomial generating function will have to skip certain powers of $q$.
Edit: Just to clarify, there is no current approach/text addressing the question:
Is there a generating function for the number of $d$-dimensional lattice paths which generate an area $t$ underneath, i.e. against the faces of the encoding cuboid?