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 $\mathcal K$?