Let $(\mathcal H_i, \xi_i), 1\leq i\leq n$ be Hilbert spaces with specified unit vectors. It follows that $$ *_{i=1}^n (B(\mathcal H_i), \xi_i) \simeq B(*_{i=1}^n (\mathcal H_i, \xi_i)) $$ where the left-hand side is the reduced free product of C$^*$-algebras and the right-hand side is the Hilbert space free product.

It struck me that I know nothing about the full (universal) free product of the $B(\mathcal H_i)$ amalgamated over $\mathbb C$. So my question is:

What is $$B(\mathcal H_1)\ \check *_\mathbb C \cdots \check *_\mathbb C \ B(\mathcal H_n)?$$ Namely, is it obviously not $*$-isomorphic to $B(\mathcal K)$ for some $\mathcal K$?

It struck me that I know nothing about the full (universal) free product of the $B(\mathcal H_i)$ amalgamated over $\mathbb C$ for Hilbert spaces $\mathcal H_i$ with identified unit vector $\xi_i$. So my question is:

What is $$B(\mathcal H_1)\ \check *_\mathbb C \cdots \check *_\mathbb C \ B(\mathcal H_n)?$$ Namely, is it obviously not $*$-isomorphic to $B(*_{i=1}^n (\mathcal H_i, \xi_i))$ or any other $B(\mathcal K)$ for some $\mathcal K$?

Edit: Thanks to Caleb Eckhardt for pointing out that $B(*_{i=1}^n (\mathcal H_i, \xi_i))$ is not $*$-isomorphic to $*_{i=1}^n (B(\mathcal H_i),\xi_i)$. Similar to the above question, I suppose one could ask whether the reduced free product is also obviously not $*$-isomorphic to some $B(\mathcal K)$.