Let $V = \mathbb{C}^n$ and let let $u$ be a $\mathbb{C}$-linear endomorphism of $V$. Then $u$ can also be considered as an $\mathbb{R}$-linear mapping $u_{\mathbb{R}}$. It is well known that $$\det u_\mathbb{R} = |\det u|^2.$$ This follows from the formula $\det u_\mathbb{R} = \mathrm{N}_{\mathbb{C}/\mathbb{R}}(\det u)$ in Bourbaki, Algèbre, III.9, Proposition 6, and the fact that $\mathrm{N}_{\mathbb{C}/\mathbb{R}}(z) = |z|^2$. In particular, this shows that $\det u_\mathbb{R} \geq 0$ (which in particular has the consequence that any complex manifold is orientable).
My question is to what extent what I've just described generalizes to the quaternions. Specifically, let $V$ now be $\mathbb{H}^n$, and let $u = u_\mathbb{H}$ be a left-$\mathbb{H}$ linear endomorphism of the vector space $V$. Since $\mathbb{C}$ is a subring of $\mathbb{H}$, we can also consider $V$ as a $\mathbb{C}$-vector space, and we can consider $u$ as a $\mathbb{C}$-linear mapping $u_{\mathbb{C}}$.
This answer shows that $\det u_{\mathbb{C}}$ is a nonnegative real number, which is analogous to the property of $u_{\mathbb{R}}$ above. But is there, analogously, a simple relationship between $\det u_{\mathbb{C}}$ and some invariant(s) of $u_\mathbb{H}$? The most likely candidate, I imagine, would be the noncommutative determinant of $u_{\mathbb{H}}$. (I know next to nothing about noncommutative determinants at the moment, but I'd still like to know if they answer my question.)
Failing that, is there a simpler proof than the one I linked to that $\det u_{\mathbb{C}}$ is a nonnegative real number?