We call a matrix good if it is symmetric positive definite with all its eigenvalues below $1$ .
Assume that we have a good matrix $B$. How to show that for $n\in \mathbb{N}$:
$$\left(\mathbb{I} - (\mathbb{I}-B)^n\right)^2\prec nB$$
I know that, given a good matrix $B$, the matrices $(\mathbb{I}-B)$ and $(\mathbb{I} - (\mathbb{I}-B)^n)$ are good as well. So, it is easy to show that $$\left(\mathbb{I} - (\mathbb{I}-B)^n\right)^2\prec I $$
But $\left(\mathbb{I} - (\mathbb{I}-B)^n\right)^2\prec nB$ does seem to be more difficult to prove.
Suppose that $Q$ is orthogonal and diagonalizes $B$. Then $$ Q'BQ = D $$ is diagonal (where the prime means 'transpose'), with all diagonal entries strictly between 0 and 1, and each column of $v$ is a unit eigenvector of $B$.
It turns out that such eigenvectors are also eigenvectors of $I - B$ (with eigenvalues $1-c$, where $c$ is the corresponding eigenvalue in $B$), hence of $(I - B)^n$, and hence of $(I - (I-B)^n)$, and hence of $(I - (I- B)^n)^2$. In short: $Q$ also diagonalizes the matrix you're looking at.
I believe (but I might be wrong -- kinda early in the morning) that in the $Q$ basis, your matrix looks like $$ (I - (I - D)^n)^2 $$ Now all you have to do is check that for each diagonal entry $c$ of $D$, you have $$ (1 - (1-c)^n)^2 \le nc $$
which is presumably either false or a not-too-hard calculus problem. I leave that part to you.
I'm a little uncertain about the less-than-or-equal vs less-than matter; I suspect that $A \preceq 0$ really means that $A$ is positive semidefinite, but I can't be sure of that. Anyhow, you can decide which one it is, and then you can do the appropriate calculus problem.