Hartree product and the Slater determinant
The Hartree product can definitely be used for approximate calculations; as you might know Hartree developed his method for atoms in 1927 (for reference, the Schrödinger equation was discovered in 1926) and he implicitely used a Hartree product for the wave function, although the connection to the exact solution of the many-body Schrödinger equation via the variational principle wasn't know. It was Slater in 1930 (and, independently, Fock) to introduce the determinant form which bears his name. When using the Hartree product one has to account for the Pauli principle "by hand", by allowing only two electrons to occupy the same spatial orbital.
By starting with a Hartree product, enforcing the Pauli principle and applying the variational principle one obtains the Hartree equations, which are similar to the Hartree-Fock ones but are missing the exchange term, usually indicated by $K$; this term is, however, smaller than the  electrostatic term $J$ so Hartree calculations are meaningful and give useful results, although of course very approximate ones.

Kohn-Sham DFT calculations use in principle the Slater product form, but in practice, depending on the exchange-correlation functional used, they lead to equations which may be more similar to the Hartree ones (i.e., without the exchange term) than to the Hartree-Fock ones. In the original LDA (local density approximation) functional the exchange term is dropped and replaced by a functional of the electron density which is much simpler to compute. As a result, DFT-LDA is similar in complexity to the Hartree method, not to the Hartree-Fock (a rather significant simplification). However, most modern functionals for molecules (as opposed to solid state calculations), e.g. B3LYP, re-introduce the exchange term (with a scaling coefficient).

The Hartree form of the wave function can be also used to compute expectation values (e.g., to compute dipole moments); you can check by looking a the Slater-Condon rules  that the results you obtain with Hartree functions are, for one-body operators, the same you'd get using the proper Slater-determinant form.