Given a symmetric non-degenerate bi-linear form $G$ on a vector space $V$, I can define the Clifford algebra from the tensor algebra $T(V)$ via:
$$ Cl(V,G) = T(V) / I_G $$ where $I_G$ is the two sided ideal generated by all elements of the form $$ I_G = v\otimes v - G(v,v). $$
Analogously if I have a symplectic form $\omega$ I may build a Weyl algebra by quotienting out the ideal generated by: $$ I_\omega = v\otimes w - w\otimes v - \omega(v,w). $$
My question whether there is an analagous algebra which can be made from some hermitian form $h$, perhaps by quotienting out by the ideal generated by $$ I_h = v^*\otimes v - h(v,v)? $$ With $h$ is conjugate linear in its first argument.
Does something like this exist and is well studied, perhaps in the Kähler theory? Or is it somehow boring, so not worth studying?
$ \newcommand\C{\mathbb C} \newcommand\Cl{\mathrm{Cl}} $Note that a conjugation $v \mapsto v^*$ is extra structure and is not inherent to having a Hermitian form.
Here is an answer that shows your proposed algebra is quite boring; I don't know if there's any way to salvage this idea and make an interesting algebra from $h$. The same reasoning ought to hold for the algebra generated by $v\otimes v - h(v,v)$ as well.
Assume $h$ is conjugate-linear in the first argument. Let its "Hermitian algebra" be denoted by $A$, identify $\C$ and $V$ as subsets of $A$ canonically, and denote the product of $A$ by juxstaposition. For any $v, w \in V$ it follows from the polarization identity $$ 4h(v,w) = h(v + w) - h(v - w) - ih(v + iw) + ih(v - iw) $$ with $h(x) = h(x,x) = x^*x$ that $$ h(v^*,w) = vw. $$ Since by definition $A$ is generated by sums of products of vectors it follows that $A = \C\oplus V$ as vector spaces, and in fact this is a $\mathbb Z/2\mathbb Z$-grading on $A$.