I am trying to prove that there is no anisotropic universal quadratic form over $\mathbb{Q}$, and here are my thoughts.
It is trivial when the dimension is $1$. When the quadratic form has dimension $2$, since it is universal, it has the form $aX_{1}^2-bX_{2}^2$, $a,b>0$ and $-a/b\in\mathbb{Q}^2$(it is anisotropic).
By the Legendre theorem and quadratic reciprocity, we can find a $t\in \mathbb{Q}$ such that $aX_{1}^2-bX_{2}^2=t$ has no rational solution, so it is not universal.
When the dimension gets larger, I think it will be easier to prove that the quadratic form cannot be anisotropic, but I can't work it out.
Any help is welcome.
ADDED: it all works. In Dickson's METN, we find, theorem 117 on page 164, that $w^2 + u^2 - 6 z^2$ is the only equivalence class of "determinant" $-6,$ attributed to Ross. Then on page 170, exercise 2, we see that $w^2 + u^2 - 6 z^2$ integrally represents all numbers other than $$ 4^k \left( 16 m + 6 \right) \; , \; \; 9^k \left( 9 m + 6 \right) \; . $$ In particular, $9-6 = 3,$ and all other primes $p \equiv 3 \pmod 8$ are so represented.
ORIGINAL:
This is provisional, I will think about it overnight. The form $$ w^2 - 2 y^2 + 3 y^2 - 6 z^2 $$ is anisotropic in $\mathbb Q_3$ three-adics. It is elementary to show that the form (integrally) represents the product of any two integers it integrally represents, that it does represent $-1,$ and that it integrally represents all (positive) primes $$ p \equiv 1,5,7 \pmod 8 $$
A computer search strongly suggests that it also represents all primes $p \equiv 3 \pmod 8.$ If that can be proved, we are done. Note tht the output below has $x=y,$ meaning that we are looking at an indefinite ternary $w^2 + v^2 - 6 z^2.$