(@NicholasTodoroff - I think that should be $(1/2)vv'$. But scale factors are irrelevant.)

The 3D case is not too difficult. Since $a\wedge b\wedge c\wedge d=0$ in fewer than four dimensions, the product $P=abcd=s+uv$ is a scalar plus a simple bivector. If the scalar $s=0$, then obviously $P=uv$ is a product of two vectors. If $s\neq0$, then we can re-scale so that $s=1$. If both $u^2=v^2=0$ then the 3D space is degenerate. So at least one of them is invertible, and $P=u(u^{-1}+v)$ is a product of two vectors.

Now to 4D. We know that a product of five vectors $P=abcde=v+T$ is a vector plus a simple trivector, and a product of three vectors also has that form. But in the latter case $v\wedge T=0$ (there can't be a $4$-vector in a 3D subspace), so we'd better have $v\wedge T=0$ in the former case also. Indeed, using the reverse $\tilde P=edcba=v-T$, we have

$$v\wedge T=\tfrac12(vT-Tv)$$ $$=\tfrac18\big((P+\tilde P)(P-\tilde P)-(P-\tilde P)(P+\tilde P)\big)$$ $$=\tfrac18\big(2\tilde PP-2P\tilde P\big)$$ $$=\tfrac18\big(2a^2b^2c^2d^2e^2-2a^2b^2c^2d^2e^2\big)=0.$$ Thus $v$ is in the subspace $T$. (I'm ignoring the trivial cases where $v=0$ or $T=0$.)

If $v$ is invertible, or if it's orthogonal to $T$ (that is, in the radical of $T$), then it can be used in an orthogonal basis: $T=vxy$. If also $x$ or $y$ is invertible, then the solution is $P=vx(x^{-1}+y)$ or $P=vy(y^{-1}-x)$. If instead $x$ and $y$ are both null, then the 4D space is degenerate.

If $v$ is not invertible and not orthogonal to $T$, then it has a complement $w\in T$ (so $v^2=w^2=0$ and $v\cdot w=1$), and there's some vector $x\in T$ orthogonal to both of them, so that $T=(v-w)(v+w)x$. Finally, using $$v=\tfrac12(v-w)(v+w)v=\tfrac12vwv,$$ we get a product of three vectors: $$P=v+T=\tfrac12(v-w)(v+w)(v+2x).$$