First, note that your edited question still has $i=2n$ when you mean $i=n$. As was pointed out by others, $n$ is the rank here and $2n$ the dimension of the first fundamental representation (the natural one for the Lie algebra in the even orthogonal case).

One useful source (if you can locate it), based on lectures given by J. Tits to mathematical physicists in Bonn many years ago, is his Springer Lecture Notes No. 40 (1967) with a German title *Tabellen zu den einfachen Lie Gruppen und ihren Darstellungen* [Tables for the simple Lie groups and their representations]. But the tables toward the end of this short volume involve little German (and Tits himself is Belgian/French). He provides explicit data including dimensions of all the fundamental representations, describing also the results for the various real forms.

In particular, the two end vertices of the Dynkin diagram usually labelled $n-1$ and $n$, correspond to the *half-spin* representations of the Lie algebra. Each has dimension $2^{n-1}$. They are not so easily constructed in terms of exterior powers of the natural representation as the earlier fundamental representations, however, with a digression into Clifford algebras usually required.

ADDED: Note that the last statement in your first paragraph is still incorrect. When you write "We know ..." this applies only for $i \leq n-2$ whereas for $n-1, n$ you get half-spin modules as discussed by Carter in his section 13.5. (This kind of concrete description of representations is found in quite a few texts.)

Concerning your questions 2, 3, it's of course possible to reach any weight vector from the highest one by subtracting various simple (or arbitrary positive) roots: By ordering the negative roots and using PBW monomials in negative root vectors, you can produce any weight space including the lowest one. The problem is that many monomials typically produce linearly dependent weight vectors, which is what makes the character theory complicated.

In particular, for $i \leq n-2$ (or for $n$ even) all fundamental representations are self-dual and therefore the lowest weight is just the negative of the highest weight under the action of the longest element $w_\circ$ of the Weyl group. (For $n$ odd, $w_\circ$ is not $-1$, so each of the last two fundamental weights goes to the negative of the other.) From the standard tables you can write down the difference between highest and lowest weights as a $\mathbb{Z}$-linear combination of simple (or arbitrary positive) roots, then construct various PBW monomials taking a highest weight vector to a lowest weight vector. Not very informative.