I am looking for an isomorphism of $\mathbb{K}$-algebras between $\Lambda(V)$ and the associated graded algebra $\mathcal{G}$ of $\mathcal{C}l(V,q) = T(V) / \mathcal{I}_q$ with $\mathcal{I}_q$ generated by $v \otimes v -q(v)1$. In my case $\mathbb{K}$ is either $\mathbb{R}$ or $\mathbb{C}$ and the quadratic form $q$ is arbitrary.
My idea goes as follows: Look at the natural filtration of the tensor algebra $T(V)$ given by $\mathcal{F}_n = \bigoplus_{k=0}^{n} V^{\otimes k}$ and project it down onto $\mathcal{C}l(V,q)$ to give a filtration $\mathfrak{F}_n = \pi_q(\mathcal{F}_n).$ Next, embed $V$ into $\mathfrak{F}_1$ and project onto $\mathfrak{F}_1 / \mathfrak{F}_0 \subseteq \mathcal{G} = \bigoplus_{n\geq 0} \mathfrak{F}_n / \mathfrak{F}_{n-1}$. This gives a linear map $\psi: V \rightarrow \mathcal{G}$ that satisfies $\psi(v)^2=0$ for all $v \in V$ because $\mathcal{G}$ is anticommutative. Now, I can use the universal property of $\Lambda(V)$ to conclude that there exists a homomorphism $\Psi$ between $\Lambda(V)$ and $\mathcal{C}l(V,q)$ satisfying $\Psi \circ i = \psi$ with the canonical injection $i: V \rightarrow \Lambda(V)$. We know that a basis for $\mathfrak{F}_n / \mathfrak{F}_{n-1}$ is given by the products $[\iota(e_{i_1}) \dots \iota(e_{i_n})]$ with strictly increasing indices between $1$ and $\dim V$ (here $\iota(v)^2=q(v)1$). The facts that $\Psi$ is a homomorphism of algebras and that $\psi$ is surjective on $\mathfrak{F}_1 / \mathfrak{F}_0$ should suffice to conclude that $\Psi$ is a surjection. I am struggling to see that it is an injection. Maybe basis elements of $\Lambda^n(V)$ are mapped to basis elements of $\mathfrak{F}_n / \mathfrak{F}_{n-1}$?
Any help or corrections would be greatly appreciated!
Over characteristic zero (say over real or complex numbers), you can present the Clifford algebra using the associated bilinear form of $q$, say $\langle u,v\rangle \colon= q(u+v) - q(u) - q(v)$. The defining relations are then that
$$u\otimes v +v\otimes u = \langle u,v\rangle$$
for $u,v\in V$. Naturally, dropping the linear term at the end defines the exterior algebra on $V$, so let us consider as you do the filtration $F$ on $\mathsf{Cliff}(V,q)$ induced from the canonical gradation on the free tensor algebra $T(V)$ where $V$ is in degree $1$.
As you noticed, since the defining relations of the exterior algebra $\Lambda(V)$ hold modulo the filtration, there is a natural map $$\psi_V: \Lambda(V) \longrightarrow \mathsf{gr}_F(\mathsf{Cliff}(V,q))$$ which you want to show is an isomorphism. It is obviously an epimorhism, since it maps onto the generators of the Clifford algebra, which generate the associated graded algebra.
To check this is an isomorphism, we can check that these algebras have the same size and, to do this, we can find a quadratic-linear Grobner basis for the Clifford algebra and compute the normal forms. The idea of the proof is completely analogous to the proof of the PBW theorem using term rewriting by Bergman, and for the "ultimate" generalization of this, you can consult say the discussion around Section 3.6.8 in the book Algebraic Operads of Loday and Vallete.
Fix a basis $(v_1,\ldots,v_n)$ of $V$, and consider the lexicographical order on $TV$ associated to $v_1<\cdots< v_n$. In this way, the defining relations of the Clifford algebra are of the form $v_iv_j = \langle v_i,v_j\rangle - v_jv_i$ for $i<j$ and $v_i^2 = \frac 12 \langle v_i,v_i\rangle$, and the normal forms (monomials in $TV$ not divisible by a leading term) are the square-free monomials in $TV$ where all the basis terms $v_i$ appear in order. In other words, normal monomials for this presentation are in bijection with the exterior algebra, as expected!
To check this is indeed a Groebner basis you need to consider the overlapping ambiguities $\underline{v_iv_j}v_k = v_i\underline{v_jv_k}$ and verify that they "resolve": if you apply the defining quadratic-linear relations in either of the underlying terms, and rewrite the resulting sum until you reach a sum of normal monomials, both results are the same.
I will do it in case $\#\{ i,j,k\} = 3$ or $2$. The case $\#\{ i,j,k\} = 1$ is very simple, as it reduces in both directions to $\frac 1 2 v_i \langle v_i,v_i\rangle$, while the case $\#\{ i,j,k\} = 2$ resolves because $\langle v,w\rangle$ is symmetric: you get
$$v_i \underline{v_i v_j} \leadsto v_i\langle v_i,v_j\rangle - \underline{v_i v_j} v_i \leadsto v_i\langle v_i,v_j\rangle - v_i\langle v_j,v_i\rangle + \underline{v_iv_i}v_j \leadsto \frac 1 2 \langle v_i,v_i\rangle v_j $$
and this is the same normal term you arrive at using $\underline{v_i v_i} v_j$. The case $\#\{ i,j,k\} = 3$ is just a bit longer, but you will arrive at $$v_i\langle v_j,v_k\rangle - \langle v_i,v_k\rangle v_j + v_k \langle v_i,v_j\rangle - v_kv_jv_i$$
and again use the the form is symmetric. The idea of this computation is checking that the relations defining $\mathsf{Cliff}(V,q)$ do not introduce new relations that may make it smaller than what "obvious" size it must have (namely, the size of the exterior algebra).