Explicit invariant of tensors nonvanishing on the diagonal The group $SL_n \times SL_n \times SL_n$ acts naturally on the vector space $\mathbb C^n \otimes \mathbb C^n \otimes \mathbb C^n$ and has a rather large ring of polynomial invariants. The element $$\sum_{i=1}^n e_i \otimes e_i \otimes e_i \in \mathbb C^n \otimes \mathbb C^n \otimes \mathbb C^n$$ is known to be GIT-semistable with respect to this action. In other words, there is a homogeneous $SL_n \times SL_n \times SL_n$-invariant polynomial, of nonzero degree, which is nonvanishing on this element. The proof I know (Theorem 4.7) uses the Hilbert-Mumford criterion and so does not explicitly construct the polynomial.

Can you give an explicit homogeneous $SL_n \times SL_n \times SL_n$-invariant polynomial that is nonzero on this element.

One can check from the definition of the hyperdeterminant that the hyperdeterminant vanishes on this element as long as $n>2$. So that won't work.
I don't know how many other natural ways there are for defining an invariant function on $\mathbb C^n \otimes \mathbb C^n \otimes \mathbb C^n$ for all $n$ simultaneously there are.
My motivation for this problem is to understand semistability of tensor powers of tensors better, and so to understand slice rank of tensor powers of tensors better. I want to give explicit criteria to show that all a tensor's tensor powers are semistable. Tensor powers of the diagonal tensor remain, so the diagonal tensor certainly has this property. Given an explicit invariant, perhaps one could find an explicit neighborhood of the diagonal tensor consisting of tensors that also have this property.
 A: Edit.  There was a mistake in the original formulation below of Theorem 2.4, p. 134 of Gelfand-Kapranov-Zelevinsky (now corrected).  That theorem only applies after passing to the standard open affines.  The upshot is that we know that there exists some integer $d_0>0$ such that the determinant $D_{A,k}^{2d}$ gives a $\textbf{SL}(A)\times \textbf{SL}(B)\times \textbf{SL}(C)$-invariant polynomial that is nonzero on Will Sawin's tensor for all $d\geq d_0$.  However, it is not immediately clear what exponent $d_0$ is necessary.  An upper bound does follow from the Castelnuovo-Mumford regularity of the symmetric power $(\mathbb{P}(A))^n//\mathfrak{S}_n$ as a subvariety of $\mathbb{P}\text{Sym}^n_k(A)$, and there are bounds on this by Gotzmann.  However, the Gotzmann bound is certainly too large for any effective computations.
Original post.
I am just writing up my comments above as an answer.  Let $k$ be a commutative, unital ring.  Let $A$ be a free $k$-module of rank $n$.  Denote by $\mathbb{P}(A)$ the Proj of the graded $k$-algebra $$S^\bullet_k(A^\vee) = \bigoplus_{d\geq 0} \text{Sym}^d_k(A^\vee).$$
For every integer $m\geq 0$, denote by $(S^\bullet_k(A^\vee))^{\otimes m}$ the $m$-fold tensor product $k$-algebra, $$(S^\bullet_k(A^\vee))^{\otimes m} = S^\bullet_k(A^\vee)\otimes_k \cdots \otimes_k S^\bullet_k(A^\vee).$$  This is naturally a $\mathbb{Z}^m$-graded algebra, and it has a natural action of the symmetric group $\mathfrak{S}_m$.  Denote by $(S^\bullet_k(A^\vee))^{\otimes m}_{\langle 1,\dots,1 \rangle}$ the $\mathbb{Z}$-graded, $\mathfrak{S}_m$-fixed subalgebra, $$S^\bullet_k(A^\vee))^{\otimes m}_{\langle 1,\dots, 1 \rangle} = \bigoplus_{d\geq 0} \text{Sym}^d_k(A^\vee)\otimes_k \cdots \otimes_k \text{Sym}^d_k(A^\vee).$$
Now assume that $m!$ is invertible in $k$, so that $\text{Sym}^m_k(A)^\vee$ is canonically isomorphic to $\text{Sym}^m_k(A^\vee)$ as $k$-vector spaces.  Then the transpose of the $m$-multilinear $k$-module homomorphism, $$A^m \to \text{Sym}^m_k(A), \ \ (a_1,\dots,a_m)\mapsto a_1\cdots a_m,$$ induces a $k$-algebra homomorphism, $$S_k^\bullet(\text{Sym}^m_k(A^\vee)) \to (S^\bullet_k(A^\vee))^{\otimes m}_{\langle 1,\dots,1 \rangle}.$$  This, in turn, defines a $k$-morphism, $$\gamma': (\mathbb{P}(A))^m\to \mathbb{P}(\text{Sym}^m_k(A)),\ \ ([a_1],\dots,[a_m]) \mapsto [a_1\cdots a_m]$$ where $(\mathbb{P}(A))^m$ is shorthand for the $m$-fold self-fiber product $\mathbb{P}(A)\times_{\text{Spec}\ k}\cdots \times_{\text{Spec}\ k}\mathbb{P}(A)$.  The morphism $\gamma'$ is invariant for the natural action of the symmetric group $\mathfrak{S}_m$ on $(\mathbb{P}(A))^m$.  Thus, $\gamma'$ factors through the quotient of this group action.  Denote by $\gamma$ the induced $k$-morphism, $$\gamma: (\mathbb{P}(A))^m//\mathfrak{S}_m \to \mathbb{P}(\text{Sym}^m_k(A)).$$
It is a classical fact that $\gamma$ is a closed immersion.  This is explicitly stated and proved in Proposition 2.1 and Theorem 2.2, Chapter 4, pp. 132-133 of Gelfand-Kapranov-Zelevinsky.  In fact, Gelfand-Kapranov-Zelevinsky prove much more: for every nonzero $x_i\in A^\vee$, denoting by $\gamma_m(x_i)\in \text{Sym}^m_k(A^\vee)$ the associated $m^{\text{th}}$ power, then the induced $k$-algebra homomorphism of affine opens, $$S_k^\bullet(\text{Sym}^m_k(A^\vee))[1/\gamma_m(x_i)] \to [(S_k^\bullet(A^\vee))^{\otimes m}_{\langle 1,\dots,1 \rangle}]^{\mathfrak{S}_m}[1/(x_{1,i}\cdots x_{m,i})],$$ is a surjective $k$-algebra homomorphism.  This is essentially Theorem 2.4 on p. 134.  From this it follows that for every $\mathfrak{S}_m$-invariant, homogeneous element $D$ of degree $r>0$ in $(S_k^\bullet(A^\vee))^{\otimes m}_{(r,\dots,r)}$, there exists an integer $e_0$ such that for every $e\geq e_0$, $D$ times the image of $\text{Sym}^e_k(\text{Sym}^m_k(A^\vee))$ is in the image of $\text{Sym}^{e+r}_k(\text{Sym}^m_k(A^\vee))$.  Thus there exists an integer $d>0$ such that $D^d$ is in the image of $\text{Sym}^{dr}_k(\text{Sym}^m_k(A^\vee))$.
In particular, for $m=n$ (assuming that $n!$ is invertible in $k$), the determinant map, $$\text{det}_{A,k}: A^n \to \bigwedge^n_k(A),$$ defines an element $D_{A,k} \in (S_k^\bullet(A^\vee))^{\otimes n}_{(1,\dots,1)}$. This element is not $\mathfrak{S}_m$-invariant.  However, the square, $D_{A,k}^2$ is $\mathfrak{S}_m$-invariant.  Thus, $D_{A,k}^{2d}$ is in the image $I_{2d}$ of $\text{Sym}_k^{2d}(\text{Sym}_k^n(A^\vee))$ for some integer $d>0$.
Finally, if we assume that $k$ is a field of characteristic $0$, then $\textbf{SL}(A)$ is linearly reductive.  Thus, the induced $k$-linear map, $$\text{Sym}^{2d}_k(\text{Sym}^m_k(A^\vee))^{\textbf{SL}(A)} \to I_{2d}^{\textbf{SL}(A)},$$
is surjective.  In particular, the $\textbf{SL}(A)$-invariant element $D_{A,k}^{2d}$ is in the image of $\left( \text{Sym}_k^{2d}(\text{Sym}_k^n(A^\vee)) \right)^{\textbf{SL}(A)}$.  So there exists some $\textbf{SL}(A)$-invariant element $\Delta_{A,k}$ mappsing to $D_{A,k}^{2d}$.
Now put this together with the tensor operation from the comments.  For $k$-vector spaces $A$, $B$, and $C$ of dimension $n$, for every $k$-linear map, $$t:A^\vee \to \text{Hom}_k(B,C),$$ there is an associated element $$\widetilde{t} \in \text{Sym}^n(A) \otimes_k \text{Hom}_k(\bigwedge^n_k(B),\bigwedge^n_k(C)),$$ giving the polynomial map, $$ A^\vee\to \text{Hom}_k(\bigwedge^n_k(B),\bigwedge^n_k(C)), \ \ a \mapsto \text{det}(t(a)).$$  The associated map $$\text{Hom}_k(A^\vee,\text{Hom}_k(B,C))\to \text{Sym}^n(A)\otimes_k \text{Hom}_k(\bigwedge^n_k(B),\bigwedge^n_k(C)), \ \ t\mapsto \widetilde{t},$$ is a degree $n$, homogeneous $k$-polynomial map that is equivariant for the natural actions of $\textbf{SL}(A)\times \textbf{SL}(B)\times \textbf{SL}(C)$.  Via the duality of $\text{Sym}^n(A)$ and $\text{Sym}^n(A^\vee)$, the polynomial $\Delta_{A,k}$ determines a $\textbf{SL}(A)\times \textbf{SL}(B)\times \textbf{SL}(C)$-invariant homogeneous polynomial of degree $2dn$ on $\text{Hom}_k(A^\vee,\text{Hom}_k(B,C))$.  
For the specific tensor $t$ that you wrote down, with respect to appropriate bases for $A$, $B$, and $C$, for every $(a_1,\dots,a_n)\in A$, $t(a_1,\dots,a_n)$ is the linear transformation whose matrix representative is the diagonal $n\times n$ matrix with entries $(a_1,\dots,a_n)$.  Thus, $\widetilde{t}$ equals $(a_1\cdots a_n)\otimes (b_1\wedge \dots \wedge b_n)\otimes (c_1\wedge \dots c_n)$.  In particular, since $a_1\cdots a_n$ is in the image of $\gamma$, $\Delta_{A,k}$ restricts on $\widetilde{t}$ as a power of the determinant.  Since $a_1\cdots a_n$ is a product of $n$ linearly independent linear polynomials, the determinant is nonzero.  Thus, $\Delta_{A,k}$ is a $\textbf{SL}(A)\times \textbf{SL}(B)\times \textbf{SL}(C)$-invariant, homogeneous polynomial of degree $2n$ that is nonzero on your tensor. 
