The notion of hyperdeterminant (in the technical sense of GKZ) is not the right one here because it is meant for non-symmetric tensors, whereas the question is about homogeneous polynomials, i.e.. symmetric tensors.

Moreover, there are two different issues to be treated separately.

If $F\in \mathbb{R}[x_1,\ldots,x_n]$ is homogeneous of degree $d$, it is not in general true that it factors as a product of linear forms $$ F(x)=\prod_{i=1}^{d}\left(\sum_{j=1}^{n}c_{ij} x_j\right) $$ even if the coefficients are allowed to be complex.

This factorization occurs iff the polynomial satisfy the Brill equations. For references on the latter see my answer to

which homogeneous polynomials split into linear factors?

Then once that is settled one needs a way to check if one can arrange for the coefficients to be real. A necessary condition is that if you restrict $F$ to a line $x=t_1 a+t_2 b$ where the $n$-component vectors $a,b$ are real, the resulting binary form in the pair of variables $(t_1,t_2)$ factorizes as a product of linear forms with real coefficients. Equivalently the associated nonhomogeneous polynomial should only have real roots, i.e., the polynomial should be real stable. See my answer to

real symmetric matrix has real eigenvalues - elementary proof

for a link to the book by Basu, Pollack and Roy on real algebraic geometry where one can find a characterization of univariate polynomials with only real roots in terms of subresultants and subdiscriminants.

I suspect the above is not only a necessary condition but also a sufficient condition, but didn't have time to think about it.

In the mentioned example, Brill equations should not be hard to write, and the real root condition is just that the discriminant of the cubic is nonnegative.

The notion of hyperdeterminant (in the technical sense of GKZ) is not the right one here, because it is meant for not necessarily symmetric tensors, whereas the question is about homogeneous polynomials, i.e., symmetric tensors.

Moreover, there are two different issues to be treated separately.

If $F\in \mathbb{R}[x_1,\ldots,x_n]$ is homogeneous of degree $d$, it is not in general true that it factors as a product of linear forms $$ F(x)=\prod_{i=1}^{d}\left(\sum_{j=1}^{n}c_{ij} x_j\right) $$ even if the coefficients are allowed to be complex.

This factorization occurs if and only if the polynomial satisfies the Brill equations. For references on the latter see my answer to

which homogeneous polynomials split into linear factors?

Then, once that is settled, one needs a way to check if one can arrange for the coefficients to be real. A necessary condition is that if you restrict $F$ to an arbitrary line $x=t_1 a+t_2 b$ where the $n$-component vectors $a,b$ are real, the resulting binary form, in the pair of variables $(t_1,t_2)$, factorizes as a product of linear forms with real coefficients. Equivalently the associated nonhomogeneous polynomial should only have real roots, i.e., the polynomial should be real stable. See my answer to

real symmetric matrix has real eigenvalues - elementary proof

for a link to the book by Basu, Pollack and Roy on real algebraic geometry where one can find a characterization of univariate polynomials with only real roots in terms of subresultants and subdiscriminants.

I suspect the above is not only a necessary condition but also a sufficient condition, but didn't have time to think about it.

In the mentioned example, Brill equations should not be hard to write, and the real root condition is just that the discriminant of the cubic is nonnegative.