Hi Darij, the main point concerning the antipode is that 

> any connected filtered bialgebra is a filtered Hopf algebra, the antipode being defined as 
> $S(x)=\sum_{k\geq0}(\eta\circ\epsilon-id)^{*k}(x)$

Here $*$ denotes the convolution product. In your example the bialgebra you consider is actually graded so the result applies. You can find the above claim (and its proof) in [these lecture notes][1] (I think you are going to like them) by Dominique Manchon (Corollary II.3.2). 

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A geometric way of seeing $T(V)$ with the shuffle product is by considering functions on the loop space of $V^*$ (i.e. the space of continuous maps from $S^1$ to $V^*$) that are given by iterated integrals. You'll see that the product of two iterated integrals is precisely the shuffle product. Moreover, this way you can see the coproduct coming from the concatenation of loops and the antipode from reversing orientation (if I remember well). 

But here you have to be in a situation when $V^{**}\cong V$


  [1]: http://arxiv.org/abs/math/0408405