I'll phrase my answer for an algebraically closed field.
As soon as all of the $n_i$ are at least 3 you get moduli. So the interesting cases are $(2,2,n)$ and $(2,3,n)$ which I claim have finitely many orbits.
Given a tensor of type $(2,2,n)$, you can always write it as $\sum e_i \otimes f_j \otimes h_{i,j}$ where $i=1,2$, $j=1,2$. Then the $h_{i,j}$ span a 4-dimensional subspace of the $n$-dimensional factor, and just using the $GL_n$ action, you can always do a transformation and assume this lies in a fixed $4$-dimensional subspace, and you're still free to use the full $GL_4$ in the stabilizer to further analyze orbit equivalence. Hence you have reduced to showing it for $(2,2,4)$. Similar remarks apply to reducing $(2,3,n)$ to $(2,3,6)$.
For $(2,2,4)$, you can replace $GL_2 \times GL_2$ acting on the $2 \times 2$ factor with $SO_4$ acting on its vector representation (you have essentially not lost any freedom by doing so). So the space you're dealing with is $4 \times 4$ matrices $V \to W$. Since one of these $4$ dimensional spaces (call it $W$) has an orthogonal form, it makes sense to take the dual to get $W \to V^\*$ (since we can identify $W = W^\*$ while respecting the group action). The two invariants that classify orbits are the rank of this matrix and the rank of the composition $V \to W \to V^*$ (left to reader). (In fact, this is true more generally whenever you consider a space of matrices between some vector space and some orthogonal (or symplectic) vector space under the action of the general linear group and the corresponding classical group.)
For $(2,3,6)$, it is more involved. There is a basic invariant, the determinant (by flattening this tensor to a $6 \times 6$ matrix). I claim that any two tensors that have nonzero determinant are in the same orbit: this is just the statement that any two invertible matrices can be transformed into one another via row operations. So now we're left with those matrices with determinant 0, which I also claim has finitely many orbits.
By a similar trick as above, we can reduce to showing that tensors of format $(2,3,5)$ have finitely many orbits. I don't think I have a direct way of seeing this, so I'll just appeal to a result of Vinberg about Z-gradings on simple Lie algebras: given a nonnegative integer valued function on the simple roots of a simple Lie algebra $\mathfrak{g}$, you can coarsen the grading given by roots by just evaluating this function on the root.
The degree 0 piece is then a reductive Lie algebra, and the result is that the simply-connected form of this Lie algebra acts on each degree $i$ piece with finitely many orbits. The case of $(2,3,5)$ comes from setting $\mathfrak{g}$ to be of type $\mathrm{E}_8$ and taking the function which just picks out the coefficient of the trivalent node of the corresponding Dynkin diagram. The degree 0 piece is $\mathfrak{sl}_2 \times \mathfrak{sl}_3 \times \mathfrak{sl}_5 \times \mathbf{C}$ and the degree 1 piece is exactly the space of $2 \times 3 \times 5$ tensors.
Last are the examples of $(2,2,2)$, $(2,2,3)$, $(2,3,3)$, and $(2,3,4)$. These also follow from Vinberg's result (pick the coefficient of the trivalent node in the Dynkin diagram of type $\mathrm{D}_4$, $\mathrm{D}_5$, $\mathrm{E}_6$, and $\mathrm{E}_7$, respectively), or maybe can be deduced from the larger examples.
This old blog post I wrote about Vinberg's result might be helpful: http://concretenonsense.wordpress.com/2010/03/01/nilpotent-orbits-in-graded-lie-algebras/
One last remark: The Vinberg result that I cited works over the complex numbers, but the finiteness of the orbits should be valid in any characteristic (in principle, the number of orbits can vary though).