Deciding bound on Tensor Rank for a fixed value This question is related to a previous unanswered question of mine.  Please see that question for background.

When Håstad proved that computing tensor rank is NP-complete, the rank in question was allowed to be part of the input.  Specifically, given a Boolean formula with $n$ variables and $m$ clauses, Håstad constructed a tensor that has rank exactly $4n+2m$ if the formula is satisfiable and has some larger rank otherwise.

What if we fix the bound on the rank to some constant?  For example...


  What is the complexity of deciding if a tensor has rank at most 3?


The tensor should be defined over an infinite field, otherwise the problem is trivial (...only a finite number of things to check).  Answers for the natural infinite fields $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ would be best.  I am personally interested in this question using the complex numbers.
Testing for constant border-rank can be done in polynomial-time, I'm pretty sure, if you're talking about tensors of fixed arity. More precisely, testing for border-rank $\leq r$ in $\mathbb{C}^{d_1} \otimes \dotsb \otimes \mathbb{C}^{d_k}$ can be done in something like $k(d_1 d_2 \dotsb d_k)^{2(r+1)}$ evaluations of $(r+1) \times (r+1)$ determinants plus some additional constant work (constant depending on $r$).

Theorem 3.7.1.1 of Landsberg's tensor book ("Inheritance") says that the equations for the variety of tensors of border-rank $\leq r$ in $\mathbb{C}^{d_1} \otimes \mathbb{C}^{d_2} \otimes \dotsb \otimes \mathbb{C}^{d_k}$ are given by the equations for border-rank $r$ in $\mathbb{C}^r \otimes \mathbb{C}^r \otimes \dotsb \otimes \mathbb{C}^r$ together with all $(r+1) \times (r+1)$ minors of flattenings. Since $r = O(1)$, there are only finitely many equations of the former type.

Border-rank 1 = rank 1, and more generally rank $\leq r$ implies border-rank $\leq r$. But border-rank 2 can have rank anywhere from $2$ to the arity $k$, and border-rank $3$ in arity 3 can have ranks 3, 4, or 5. Beyond that I don't know about testing for rank.