$\newcommand{\qalg}{\mathcal{I}_q(V)}$As K. Conrad points out, this question has actually been answered already on MathOverflow by user MTS: see these answers here and here.
The essential idea is this: despite the fact that Clifford algebras have non-zero quadratic form in general, we can still "piggyback" on the embedding of the exterior algebra in the tensor algebra to represent the Clifford algebra, regardless of the choice of quadratic form.
Note: The embedding of the exterior algebra into the tensor algebra can be given by defining the exterior=wedge product in terms of the tensor product. Discussion of two canonical ways of doing this are given here on MathOverflow (the top answer being again by MTS) as well as on Math.SE.
Beautiful and rigorous proofs of this fact are given in the two answers (1) (2) linked to above by user MTS; in what follows I will merely try to motivate this result for novices like me.
1. First, we might expect this result intuitively by considering the decomposition of the Clifford product for vectors into an "inner" and "outer" product. The former is the inner product generated by polarization of the quadratic form $q$ and the latter is "equivalent" in some sense to the exterior/wedge product from the exterior algebra. Since the inner product is grade-reducing, while the outer product is grade-increasing, we might expect that the inner product would be "incapable of producing new basis elements", hence the only basis elements of the Clifford algebra we should expect are those created by the corresponding exterior algebra, i.e. there is some reason to expect a priori that a given Clifford algebra is "no more rich than" making the corresponding exterior algebra an inner product space, i.e. by defining a quadratic form on the exterior algebra.
2. MTS's answers make this suspicion rigorous by showing that the ideals for the Clifford algebra and the exterior algebra are sufficiently similar so that the quotients of the tensor algebra by them can be related in a straightforward manner.
Let us motivate this fact by considering $\mathcal{I}_q(V)$ explicitly. Remember that $$\mathcal{I}_q(V):= \left\{ \sum_k A_k \otimes (v \otimes v - q(v)) \otimes B_k : v \in V, A_k, B_k \in \mathcal{T}(V) \right\}$$ Using the distributivity of the tensor product (e.g. here) I rewrite this as $$\qalg = \left\{ \sum_k \left[A_k \otimes v \otimes v \otimes B_k - A_k \otimes q(V) \otimes B_k \right] : v \in V, A_k, B_k \in \mathcal{T}(V)\right\} \\ \overset{?}{=} \left\{ \sum_k \left[A_k \otimes v \otimes v \otimes B_k - q(V) \cdot( A_k \otimes B_k) \right] : v \in V, A_k, B_k \in \mathcal{T}(V)\right\}$$
In any case, the motivating idea here is that the right term arising from the quadratic form, always being of strictly lower grade than the left term, should in some sense be "negligible" compared to the left term (MTS's second answer makes this idea explicit and rigorous). In other words: $$\qalg \approx \left\{ \sum_k A_k \otimes v \otimes v \otimes B_k: v \in V, A_k, B_k \in \mathcal{T}(V) \right\}$$ where the right hand side is obviously $\mathcal{I}_0(V)$, the ideal which forms the exterior algebra.
3. In the very simple case of geometric algebra over $\mathbb{R}^n$, this fact is used all the time in creating a basis for the geometric algebra. Namely, it is stated that $\mathbb{G}^n \cong \mathbb{R}^{2^n}$ (vector space isomorphism), with the basis being given by: $$\{e_1, \dots, e_n, e_{\sigma_2(1)}\wedge e_{\sigma_2(2)}, e_{\sigma_3(1)}\wedge e_{\sigma_3(2)} \wedge e_{\sigma_3(3)}, \dots, e_1 \wedge \dots \wedge e_n : \sigma_i \in A_i \}$$ where $A_i$ denotes the set of all even permutations of $i$ elements selected from $\{1, \dots, n\}$ and $\{e_1,\dots,e_n\}$ is an orthonormal basis of $\mathbb{R}^n$, orthogonal with respect to the inner product of $\mathbb{G}^n$.
It should be evident that this basis is isomorphic to the basis of the exterior algebra over $\mathbb{R}^n$.
Since in any Clifford algebra $vw = v \wedge w$ ($vw$ denoting the Clifford/geometric product) if and only if $\langle v, w \rangle_q=0$ i.e. if and only if $v$ and $w$ are orthogonal with respect to the inner product induced by $q$ via polarization, the possible choices of orthonormal basis will in general depend on the choice of quadratic form for the geometric algebra. Nevertheless, the one-to-one correspondence of the bases with the basis for the exterior algebra $\mathbb{R}^n$ will still remain and will be sufficient to generate a vector space isomorphism, since vector space isomorphisms only depend on the linear independence and dimensions of the bases, and not their orthogonality under an inner product.
In other words, we can always use the vector space basis of the exterior algebra over $\mathbb{R}^n$ as a vector space basis for the geometric algebra $\mathbb{G}^n$ regardless of our choice of quadratic form, although this basis will not always be orthogonal. Hence by identifying the geometric algebra with the exterior algebra + inner product structure, we get a linear embedding of the geometric algebra into the tensor algebra "for free" by using the exterior algebra's embedding.
(See p.4 of this document for discussion of the canonical basis of a geometric algebra generated by an orthonormal basis of $\mathbb{R}^n$.)
Best Answer
Non-homogeneous Koszul duality is now well-understood. Here are a few references:
As far as I remember the new book of Loday and Vallette discusses this too (see $\S 3.6$).
You can find the statement that Weyl and Clifford algebras are Koszul in the inhomogenous sens in this paper of Braverman-Gaistgory ($\S 5.3$).
Nevertheless, as it is said in Leonid Positselski's comment, Weyl and Clifford algebras are not Koszul dual to each other. The reason is that inhomogeneous Koszul duality is inhomogeneous!
quadratic-linear algebras are dual to DG quadratic algebras (e.g. the universal envelopping algebra of a Lie algebra is Koszul dual its Chevalley-Eilenberg algebra).
quadratic--linear-constant algebra (e.g. Weyl or Clifford, for which there is even no linear part) are dual to curved quadratic DG algebras. E.g. for the Weyl algebra $\mathcal W_{(V,\omega)}$, its Kozsul dual is the pair $(\wedge(V^*),\omega)$ where the symplectic form $\omega$ is viewed as a curvature (a degree 2 element) in the exterior algebra.