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).

Edit: I forgot about $(2,4,4)$. Such a tensor can be interpreted as a $4 \times 4$ matrix of linear forms in two variables $x,y$, and hence their determinant is a quartic binary form (and you can get all such in this way by putting the irreducible factors as the diagonal entries). These have moduli (for instance, generic quartic binary forms produce elliptic curves by taking the associated double cover of the projective line). Any other format not discussed either has all $n_i \ge 3$ or at least two $n_i \ge 4$ and hence has moduli since it contains a case that we already discussed.