Well, unless I am mistaken, whenever you have a cdg ring given by dg generators and relations, then the cotangent space at zero is simply the cone of the differential $T^*\text{Spec}(S^*{\text{generators}})\to T^*\text{Spec}(S^*{\text{relations}})$ at the corresponding point. In this case, as $d(\text{anything quadratic}) = 0$ at $0\in V$, this should give a complex with zero differential, $V\oplus S^2(V)[-1]$ at the point (here $S^2(V^*)[1]$ means it's in degree $1$, not a shift).

More concretely in this case, you have a Koszul CDG-free resolution for $X$ as follows: $A = \left(\bigwedge^* S^2(V)\otimes S^*[V], d\right)$ with $S^2(V)$ in degree -1, with differential determined on the degree-minus-one generators by $d(v\otimes w) = vw\in S^*(V)$. It's more intuitive for me to understand the dual problem of computing the tangent rather than the cotangent complex, i.e. derivations $A\to \mathbb{C}$. As a graded space, this is dual to the space of generators, so $V^*\oplus S^2(V^*)[-1]$, where the differential $d:V^*\to S^2(V^*)$ is $d(\partial_\xi) = [d, \partial_\chi] = -\partial_\xi\circ d_A$ acts on $v w\in S^2(V)$ by $d(\partial_\xi)(vw[-1]) = \partial_\xi(vw[0]) = v\partial_xi(w) \pm w \partial_\xi(v)$ (here I'm using the crude notation $[0], [-1]$ to distinguish degree 1 and degree 0 elements), and as both $\partial_\xi(v), \partial_\xi(w)\in \mathbb{C}$ are killed by $V$, this just gives us $d=0,$ in the tangent complex hence the desired result.