A question about the Tannakian etale fundamental group of a curve

Let $X$ be a smooth connected quasi-projective curve over $\mathbf{Q}$. Let $U$ be the pro-unipotent etale fundamental group of $X$ over $\mathbf{Q}_p$.

$U^1 = U$ and let $U^n =[U,U^{n-1}]$.

Let $n\geq 1$.

Why is $U^n/U^{n+1} \cong \mathbf{Q}_p(n)^{r_n}$ for some positive integer $r_n$?

I "know" this is true for $X=\mathbf{P}^1-\{0,1,\infty\}$ because M. Kim uses it in his article on Siegel's theorem and the motivic fundamental group.

Why is this true in general? (Here we should probably ask our curve to be hyperbolic.)

Is this true if we replace $X$ by a higher-dimensional variety in general?

Does the above property also hold for the "other" unipotent fundamental groups such as the pro-unipotent de Rham fundamental group of $X$? (Again, the answer is yes if $X$ is the projective line minus three points.)

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This is not true for curves whose compactification has genus $>0$. – ulrich Apr 28 '12 at 8:57

Al ulrich says, this is not true in general. If $X$ is non-compact (i.e. affine), then the Lie algebra of U is isomorphic to the free Lie algebra on $H^1_\mathrm{et}(X,\mathbb{Q}_p)^\vee$. Hence in the case of $\mathbb{P}^1\setminus \{0,1,\infty \}$, where the first etale cohomology is $\mathbb{Q}_p(-1)^{\oplus 2}$, it follows that $U^n/U^{n+1}$ is isomorphic to $\mathbb{Q}_p(n)^{r_n}$ where $r_n$ is the dimension of the $n$th graded part of the free Lie algebra on a 2-dimensional vector space, but for $X$ with a higher genus compactification, what you claim will fail even for $U^1/U^2\cong H^1_\mathrm{et}(X,\mathbb{Q}_p)^\vee$.