The operator-norm ball is the convex hull of O(n)

Idea. The singular value decomposition reduces the problem to diagonal matrices with entries in $[-1,1]$, and the cube $[-1,1]^n$ is the convex hull of its vertices $\{\pm1\}^n$.

Step 1: $\operatorname{conv}O(n)\subseteq\mathcal B$. Every orthogonal matrix has norm $1$, so $O(n)\subset\mathcal B$. The unit ball of any norm is convex, hence contains the convex hull of $O(n)$.

Step 2: diagonal matrices. Let $\sigma\in[0,1]^n$. Put $\lambda_i=\frac{1+\sigma_i}2\in[\frac12,1]$, so that $\sigma_i=\lambda_i\cdot(+1)+(1-\lambda_i)\cdot(-1)$. For each sign vector $\varepsilon\in\{\pm1\}^n$ define the weight
$$w_\varepsilon=\prod_{i=1}^n\begin{cases}\lambda_i&\text{if }\varepsilon_i=+1\\1-\lambda_i&\text{if }\varepsilon_i=-1.\end{cases}$$
Then $w_\varepsilon\ge0$, and $\sum_\varepsilon w_\varepsilon=\prod_i(\lambda_i+1-\lambda_i)=1$. Moreover the $i$-th diagonal entry of $\sum_\varepsilon w_\varepsilon\operatorname{diag}(\varepsilon)$ is $\sum_\varepsilon w_\varepsilon\varepsilon_i=\lambda_i-(1-\lambda_i)=\sigma_i$ (the weights of the other coordinates sum to $1$). So
$$\operatorname{diag}(\sigma)=\sum_{\varepsilon}w_\varepsilon\operatorname{diag}(\varepsilon).$$

Step 3: general case. Let $A\in\mathcal B$ with SVD $A=P\operatorname{diag}(\sigma)Q^T$, $\sigma_i\in[0,1]$. Then
$$A=\sum_\varepsilon w_\varepsilon\,P\operatorname{diag}(\varepsilon)Q^T,$$
a convex combination of the orthogonal matrices $P\operatorname{diag}(\varepsilon)Q^T$.

Probabilistic reading. If $\varepsilon_i$ are independent random signs with $\mathbb{P}(\varepsilon_i=1)=\lambda_i$, then $\mathbb{E}[\operatorname{diag}(\varepsilon)]=\operatorname{diag}(\sigma)$.

Remark. The hull of $O(n)$ is compact (the convex hull of a compact set in finite dimension is compact), so it is automatically *closed*, consistent with the statement. Note that the convex hull of $SO(n)$ is strictly smaller (every orthogonal matrix of determinant $-1$ is an extreme point of $\mathcal B$ by the previous problem, so it cannot be a convex combination of elements of $SO(n)$ other than itself), and the representation above uses $2^n$ orthogonal matrices; by Carathéodory's theorem a representation with at most $n^2+1$ matrices exists.