Clifford algebras are superalgebras — i.e., they are $\mathbb{Z}_2$-graded — because they can be expressed as a quotient of the tensor algebra by an ideal which is $\mathbb{Z}_2$-graded.
There is a theorem which states a graded ideal results in a graded quotient algebra.
These main points are summarised below.
Def. An algebra $A$ is $R$-graded for a monoid $R$ (e.g., $\mathbb{Z}$ or $\mathbb{Z}_2$) if there is a decomposition
$$
A = \bigoplus_{k \in R} A_k
$$
which preserves multiplication,
$$
A_p A_q \subseteq A_{p + q}
\qquad\text{i.e.,}\qquad
a \in A_p \text{ and }b \in A_q \implies ab \in A_{p + q}
.$$
For example, the tensor algebra $V^\otimes = \bigoplus_{k=0}^\infty {V_{(1)} \otimes \cdots \otimes V_{(k)}}$ is $\mathbb{Z}$-graded, since the tensor product of rank $p$ and $q$ tensors is rank $p + q$. In fact, $V^\otimes$ is $\mathbb{Z}_n$-graded for any $n$.
Def. An ideal $I \subseteq A$ of an $R$-graded algebra $A$ is itself $R$-graded (a.k.a. homogeneous) if
$$
I = \bigoplus_{k \in R} I_k
\quad\text{where}\quad
I_k = I \cap A_k
.$$
For example, the ideal $I_{alt}$ of $V^\otimes$ generated by $\{a\otimes b + b\otimes a\}$ is $\mathbb{Z}$-graded, because $I_{alt} = I_{alt} \cap (V\otimes V)$.
Thm. If $A$ is an $R$-graded algebra, and $I$ is an $R$-graded ideal, then the quotient algebra $A/I$ is also $R$-graded.
For example, this implies $V^\otimes \big/ I_{alt} \cong \bigwedge V$ is $\mathbb{Z}$-graded.
On the other hand, a Clifford algebra is a quotient
$$
Cl(V, Q) = V^\otimes \big/ \langle a \otimes a - Q(a) \rangle
$$
by the ideal generated by $a\otimes a \sim Q(a)$.
This ideal is not $\mathbb{Z}$-graded. However, it is $\mathbb{Z}_2$-graded (it is generated by the sum of a scalar and rank-$2$ tensor).
Since $V^\otimes$ is also $\mathbb{Z}_2$-graded, the theorem above shows that $Cl(V, Q)$ is itself $\mathbb{Z}_2$-graded (a.k.a., a superalgebra).
Proofs can be found e.g., in my unpublished thesis, where I copied these from.
