For simplicity, work over $\mathbb F_2$ and only consider order-$3$ equilateral tensors. For $r\in\mathbb Z_{>0}$, how many tensors $\mathfrak{T}\in\mathsf{Ten}_{n}^{\otimes 3}(\mathbb F_2)$ are there of tensor rank $r$? ($\mathfrak{T}$ is Gothic $T$.) Here $\mathsf{Ten}_n^{\otimes 3}(\mathbb F_2)$ denotes the set of all $n\times n\times n$ tensors over $\mathbb F_2$. In general, the tensor rank of a tensor $\mathfrak{T}\in\mathsf{Ten}_{\ell,m,n}(\mathbb F)$ is defined as $$\mathrm{trk}(\mathfrak{T})=\min\left\{r\in\mathbb Z_{>0}\colon\mathfrak{T}=\sum_{i=1}^r\vec{u}\otimes\vec{v}\otimes \vec{w} \text{ for }\vec u\in\mathbb F^\ell,\vec v\in\mathbb F^m,\vec w\in\mathbb F^n\right\}.$$

The same question is easy for matrices over $\mathbb F_2$ (and indeed over finite field of any characteristic). Note that picking an $r$-dimensional subspace of $\mathbb F_2^n$ can be thought as picking a rank $r$ matrix of size $n\times k$ for some $n\ge k\ge r$ (for the current purpose, take $k=n$), modulo picking a $k\times r$ matrix of rank $r$. $$\# r\text{-dim subspaces}=\frac{\#n\times k\text{ matrices of rank }r}{\#k\times r\text{ matrices of rank }r}.$$

The number of $r$-dimensional subspaces is given by the famous Gaussian binomial coefficient $\begin{bmatrix}n\\r\end{bmatrix}_2$. The number of full column-rank $k\times r$ matrix ($k>r$) can be obtained by sequentially picking its columns so that the $i$-th column is linearly independent of the first $i-1$ columns. This gives $\prod_{i=0}^{r-1}(2^k-2^i)$. Hence the number of rank-$r$ $n\times k$ matrices is $$\begin{bmatrix}n\\r\end{bmatrix}_2\cdot \prod_{i=0}^{r-1}(2^k-2^i)=\prod_{i=0}^{r-1}\frac{(2^n-2^i)(2^k-2^i)}{2^r-2^i}.$$

I do not see how to apply such an idea to tensors. For one thing, we can think a tensor as generating a vector space of matrices. Some people call it a *slice space* of the tensor (cf. page 2 of this paper). However, there does not seem to be a higher order notion of Gaussian coefficient for a subspace of $\mathsf{Mat}_{n\times n}(\mathbb F_2)$. For another thing, tensors have different directions. By that I mean, an order-$3$ tensor has three directions. We cannot pick sequentially two-dimensional slices, i.e., three $n\times n$ matrices, and claim that the tensor rank is bounded desiredly.

My question is: how can one count the number of $\ell\times m\times n$ tensors over $\mathbb F_q$ of rank $r$?