According to wikipedia, a clifford algebra is the freest algebra over a vector space $V$ equipped with a quadradic form $Q:V \to F$ where $F$ is a field. For instance $1Q(v)=v^2$ forall $v \in V$.
Does the quadratic form apply only to 1-vector, or to all muli-vector? What grades of vectors does $V$ contain?
For instance, does $Q$ maps $v=1+2e_0+4e_2e_3$ to a real number?
If so, how do I calculate it?
By definition, $V$ is the space of $1$-vectors, and $Q$ applies to $1$-vectors.
But there is a natural extension of $Q$ to the whole algebra; see my answer to Inner product structure on geometric algebra?. Applied to your example, it is simply
$$Q'(1+2e_0+4e_2e_3)=1^2+2^2+4^2=21,$$
assuming the Euclidean form $Q(\alpha_0e_0+\alpha_1e_1+\alpha_2e_2+\alpha_3e_3)=\alpha_0\!^2+\alpha_1\!^2+\alpha_2\!^2+\alpha_3\!^2$. If instead you use the Lorentzian form $Q(\alpha_0e_0+\alpha_1e_1+\alpha_2e_2+\alpha_3e_3)=-\alpha_0\!^2+\alpha_1\!^2+\alpha_2\!^2+\alpha_3\!^2$, then
$$Q'(1+2e_0+4e_2e_3)=1^2-2^2+4^2=13.$$
