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What follows is incorrect, as pointed out by Darij and Theo. In particular, it doesn't seem that anywhere I am using the fact that $g$ is a quadratic form. I am leaving it here in the hope that it can be somehow rescued by additional assumptions, or some adaptation of the method will work.

I believe this follows from the general statement that taking the associated graded object is an exact functor from the category of filtered $k$-algebras to the category of graded $k$-algebras. Denote the tensor algebra of $L$ by $T(L)$, denote the relevant ideal by $$I = \langle x \otimes x - g(x) \rangle$$ and the inclusion by $\iota: I \to T(L)$.

Then the morphism $\mathrm{gr} (\iota)$ of graded $k$-algebras maps $x \otimes x - g(x)$ (or, more pedantically, its image in $\mathrm{gr}(I)$) to $x \otimes x$ (after identifying $\mathrm{gr}(T(L))$ with $T(L)$).

This shows that the associated graded algebra of $\mathrm{Cl}(L,g)$ is $\bigwedge(L)$, and this lifts to an isomorphism of $k$-modules between $\mathrm{Cl}(L,g)$ and $\bigwedge(L)$.

Edit: I have expanded my original answer.

Let $A = \cup_{n=0}^\infty A_n$ be any filtered $k$-algebra and let $I$ be an ideal. Denote $I_n = I \cap A_n$, so $I$ is filtered by $I = \cup_{n=0}^\infty I_n$. Let $$q: A \to A/I$$ be the quotient map. Then $A/I$ is filtered by $$A/I = q(A) = \cup_{n=0}^\infty q(A_n).$$ Let the inclusion map be denoted $\iota : I \to A$. Then we have the short exact sequence $$0 \to I \overset{\iota}{\longrightarrow} A \overset{q}{\longrightarrow} A/I \to 0$$ in the category of filtered $k$-algebras.

My claim is that the sequence $$0 \to \mathrm{gr}(I) \overset{\mathrm{gr}(\iota)}{\longrightarrow} \mathrm{gr}(A) \overset{\mathrm{gr}(q)}{\longrightarrow} \mathrm{gr}(A/I) \to 0$$ remains exact in the category of graded $k$-algebras. This is a bit tedious to check but not difficult.

Then apply this to $A = T(L)$, $I = \langle x \otimes x - g(x) \rangle$. Then $A/I$ is your Clifford algebra, and I am saying that you can figure out its associated graded object as $$\mathrm{gr}(A/I) \simeq \mathrm{gr}(A)/ \mathrm{gr}(\iota)(\mathrm{gr}(I)).$$ Finally, what does the map $\mathrm{gr}(\iota)$ do to the ideal? It turns out that $$\mathrm{gr}(\iota) (x \otimes x - g(x)) = x \otimes x,$$ so the image of $\mathrm{gr}(I)$ under $\mathrm{gr}(\iota)$ is exactly the ideal you quotient by to get the exterior algebra $\bigwedge(L)$.

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Edit: I have expanded my original answer.

Let $A = \cup_{n=0}^\infty A_n$ be any filtered $k$-algebra and let $I$ be an ideal. Denote $I_n = I \cap A_n$, so $I$ is filtered by $I = \cup_{n=0}^\infty I_n$. Let$$q: A \to A/I$$be the quotient map. Then $A/I$ is filtered by $$A/I = q(A) = \cup_{n=0}^\infty q(A_n).$$Let the inclusion map be denoted $\iota : I \to A$. Then we have the short exact sequence $$0 \to I \overset{\iota}{\longrightarrow} A \overset{q}{\longrightarrow} A/I \to 0$$in the category of filtered $k$-algebras.

My claim is that the sequence$$0 \to \mathrm{gr}(I) \overset{\mathrm{gr}(\iota)}{\longrightarrow} \mathrm{gr}(A) \overset{\mathrm{gr}(q)}{\longrightarrow} \mathrm{gr}(A/I) \to 0$$remains exact in the category of graded $k$-algebras. This is a bit tedious to check but not difficult.

Then apply this to $A = T(L)$, $I = \langle x \otimes x - g(x) \rangle$. Then $A/I$ is your Clifford algebra, and I am saying that you can figure out its associated graded object as$$\mathrm{gr}(A/I) \simeq \mathrm{gr}(A)/ \mathrm{gr}(\iota)(\mathrm{gr}(I)).$$Finally, what does the map $\mathrm{gr}(\iota)$ do to the ideal? It turns out that$$\mathrm{gr}(\iota) (x \otimes x - g(x)) = x \otimes x,$$so the image of $\mathrm{gr}(I)$ under $\mathrm{gr}(\iota)$ is exactly the ideal you quotient by to get the exterior algebra $\bigwedge(L)$.

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I believe this follows from the general statement that taking the associated graded object is an exact functor from the category of filtered $k$-algebras to the category of graded $k$-algebras. Denote the tensor algebra of $L$ by $T(L)$, denote the relevant ideal by $$I = \langle x \otimes x - g(x) \rangle$$ and the inclusion by $\iota: I \to T(L)$.

Then the morphism $\mathrm{gr} (\iota)$ of graded $k$-algebras maps $x \otimes x - g(x)$ (or, more pedantically, its image in $\mathrm{gr}(I)$) to $x \otimes x$ (after identifying $\mathrm{gr}(T(L))$ with $T(L)$).

This shows that the associated graded algebra of $\mathrm{Cl}(L,g)$ is $\bigwedge(L)$, and this lifts to an isomorphism of $k$-modules between $\mathrm{Cl}(L,g)$ and $\bigwedge(L)$.