I was reading the section on Clifford algebras in Algebra by Serge Lang (one of my favourite authors), when I stumble right at the start of his exposition!
Notation: I borrow also from Bourbaki in addition to Lang. Let $A$ be a commutative ring and $E$ an $A$-module (in Lang $A = k$ is a field, but Bourbaki has it more general). Further let $Q$ be a quadratic form (Lang instead chooses $g$ a symmetric bilinear form everything is the same as long as we use asterisks).
Lang defines the Clifford algebra $C(Q)$ (alternatively $C_Q(E)$) corresponding to $(E,Q)$ to be an $A$-algebra together with a linear map $\rho:E \rightarrow C(Q)$, such that $C(Q)$ solves the following universal problem: if $\psi: E \rightarrow L$ be a linear map into any $A$-algebra $L$ such that $\psi(x)^2 = Q(x) \cdot 1$ (where $1$ is the unit element of $L$), then there exists a unique algebra homomorphism $C(\psi) = \psi_\star: C(Q) \rightarrow L$ such that the relevant triangle diagram commutes (i.e. $\psi_* \circ \rho = \psi$).
He then writes by abstract nonsense, $C(Q)$ is uniquely determined up to isomorphism and we have trivially $\rho(x)^2 = Q(x)\cdot 1$.
Maybe I am being very slow today, but surely he means to write $\rho(x)^2 = Q(x)\cdot 1$ in the definition of $C(Q)$. How can this possibly follow from his definition?
You are morally right, it should be included in the definition.
This being said, it is not formally needed, it turns out it does follow from the given definition. Indeed, if we "cheat" a little bit, we actually know that there is a Clifford algebra $L$, with a canonical map $\psi: E\to L$ which is injective. Then using the universal property we get $\psi_*: C(Q)\to L$, and $\psi_*\circ \rho$ must be injective, so from $$\psi(x)^2 = Q(x)\cdot 1_L$$ we get $$\psi_*(\rho(x)^2) = Q(x)\cdot 1_L$$ and by injectivity $$\rho(x)^2 = Q(x)\cdot 1_{C(Q)}.$$
But this trick depends on the fact that there exists some $(L,\psi)$ with $\psi$ injective, so essentially on the explicit construction of the Clifford algebra itself. I agree that we should instead require that $\rho(x)^2=Q(x)\cdot 1_{C(Q)}$ in the universal property.