Consider the projective quadric $V$ given by $$ 2x^2+y^2+z^2+w^2=0 $$ over $\mathbb{Q}$. Inspired by Jason Starr's remark on splitting fields, I will prove that $V$ is not birational to a product of Severi-Brauer varieties over $\mathbb{Q}$.
Clearly $V$ has no points over $\mathbb{R}$, hence no rational points. So if $V$ is birational to a product $\prod_i W_i$ of Severi-Brauer varieties, then we must have $\dim W_i = 1$ for all $i$ since the $W_i$ have points over a quadratic extension but not all of them have points over $\mathbb{Q}$. Hence $V \sim W_1 \times W_2$. Also, $V$ has points over $\mathbb{Q}_p$ for $2 < p < \infty$ by Chevalley-Warning and Hensel's lemma. So both $W_i$ have points over $\mathbb{Q}_p$ for $2 < p < \infty$, which implies that the $W_i$ are isomorphic to either $\mathbb{P}^1$ or to $W':x^2+y^2+z^2=0$, since these are the only conics over $\mathbb{Q}$ up to isomorphism that have points over $\mathbb{Q}_p$ for all $2 < p < \infty$; moreover, not both $W_i$ can be isomorphic to $\mathbb{P}^1$. This then implies that $W_1 \times W_2$ has no points over $\mathbb{Q}_2$, while $V$ contains the point $(1,1,2,\sqrt{-7})$ which is defined over $\mathbb{Q}_2$.
Edit: It might be enlightening to generalize the example a little bit. Assume that $V$ is a smooth quadric over a number field $k$ that is birationally equivalent to a finite product $\prod_i W_i$ of Severi-Brauer varieties. If we define $S$ as the set of places $v$ of $k$ such that $V(k_v) = \emptyset$ and for each $i$ we define $S_i$ as the set of places such that $W_i(k_v) = \emptyset$, then by the Lang-Nishimura lemma we have $$ S = \bigcup_i S_i. $$ By the reciprocity law for the Brauer group, each $S_i$ has an even number of elements. It follows that $S$ cannot consist of a single place. This explains the example above, where $V(\mathbb{Q}_p) \neq \emptyset$ for all primes $p$, but $V(\mathbb{R}) = \emptyset$; hence $\# S=1$ and there cannot exist $W_i$ such that $V \sim \prod_i W_i$.