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I think that the part $(a)$ of proposition $2.4$ of this paper shows that for $n$ sufficiently large one can construct a $n \times n$ unitary matrix $U=-I_{2}\oplus U'$, which can be decomposed as the product of three positive matrices. For such $U$ we have $\parallel U-I\parallel=2$. In fact one can take a unitary matrix $U'$ with $Det \;U'=1$ such that $0$ lies in the interior of the convex hull of the eigenvalues of $U$. This convex hull is equal to $W(U)$, the numerical range of $U$.

In fact if $U=ABC$ for positive $A,B,C$ then $UC^{-1}=AB$ so $C^{-1}=|AB|$ is the positive factor of $U $ in its polar decomposition. Of course such $U$ has $-1$ as an eigenvalue.

I think that the part $(a)$ of proposition $2.4$ of this paper shows that for $n$ sufficiently large one can construct a $n \times n$ unitary matrix $U=-I_{2}\oplus U'$, which can be decomposed as the product of three positive matrices. For such $U$ we have $\parallel U-I\parallel=2$. In fact one can take a unitary matrix $U'$ with $Det \;U'=1$ such that $0$ lies in the interior of the convex hull of the eigenvalues of $U$. This convex hull is equal to $W(U)$, the numerical range of $U$.

In fact if $U=ABC$ for positive $A,B,C$ then $UC^{-1}=AB$ so $C^{-1}=|AB|$ is the positive factor of $U $ in its polar decomposition. Of course such $U$ has $-1$ as an eigenvalue.

An explicit example is the following

$$\begin{pmatrix}-1& 0&0&0\\0&-1&0&0\\0&0&\sqrt{3}/2&-1/2\\0&0& 1/2&\sqrt{3}/2 \end{pmatrix} $$ According to the aboved linked paper this matrix is a product of three positive matrices. This matrix, with distance $2$ from the identity matrix $I_{4}$, is the unitary factor of polar decomposition of $AB$ for two positive matrices $A,B$. So according to the notation $C_{n}$ in the answer by Loup Blanc, we have $C_{n}=2, \; \forall n\geq 4$.

I think that the part $(a)$ of proposition $2.4$ of this paper shows that for $n$ sufficiently large one can construct a $n \times n$ unitary matrix $U=-I_{2}\oplus U'$, which can be decomposed as the product of three positive matrices. For such $U$ we have $\parallel U-I\parallel=2$. In fact one can take a unitary matrix $U'$ with $Det \;U'=1$ such that $0$ lies in the interior of the convex hull of the eigenvalues of $U$.

In fact if $U=ABC$ for positive $A,B,C$ then $UC^{-1}=AB$ so $C^{-1}=|AB|$ is the positive factor of $U $ in its polar decomposition. Of course such $U$ has $-1$ as an eigenvalue.

I think that the part $(a)$ of proposition $2.4$ of this paper shows that for $n$ sufficiently large one can construct a $n \times n$ unitary matrix $U=-I_{2}\oplus U'$, which can be decomposed as the product of three positive matrices. For such $U$ we have $\parallel U-I\parallel=2$. In fact one can take a unitary matrix $U'$ with $Det \;U'=1$ such that $0$ lies in the interior of the convex hull of the eigenvalues of $U$. This convex hull is equal to $W(U)$, the numerical range of $U$.

In fact if $U=ABC$ for positive $A,B,C$ then $UC^{-1}=AB$ so $C^{-1}=|AB|$ is the positive factor of $U $ in its polar decomposition. Of course such $U$ has $-1$ as an eigenvalue.

I think that the part $(a)$ of proposition $2.4$ of this paper shows that for $n$ sufficiently large one can construct a unitary matrix $U=-I_{2}\oplus U'$, which can be decomposed as the product of three positive matrices. For such $U$ we have $\parallel U-I\parallel=2$. In fact one can take a unitary matrix $U'$ with $Det \;U'=1$ such that $0$ lies in the interior of the convex hull of the eigenvalues of $U$.

In fact if $U=ABC$ for positive $A,B,C$ then $UC^{-1}=AB$ so $C^{-1}=|AB|$ is the positive factor of $U $ in its polar decomposition. Of course such $U$ has $-1$ as an eigenvalue.

I think that the part $(a)$ of proposition $2.4$ of this paper shows that for $n$ sufficiently large one can construct a $n \times n$ unitary matrix $U=-I_{2}\oplus U'$, which can be decomposed as the product of three positive matrices. For such $U$ we have $\parallel U-I\parallel=2$. In fact one can take a unitary matrix $U'$ with $Det \;U'=1$ such that $0$ lies in the interior of the convex hull of the eigenvalues of $U$.

In fact if $U=ABC$ for positive $A,B,C$ then $UC^{-1}=AB$ so $C^{-1}=|AB|$ is the positive factor of $U $ in its polar decomposition. Of course such $U$ has $-1$ as an eigenvalue.