The question is not well-formulated, as Bugs Bunny observes. In the case of the special linear group or Lie algebra (say over the field $\mathbb{C}$), it happens to be true that the fundamental dominant weights are precisely the minuscule ones (defined by the condition that the weight is nonzero while all pairings with arbitrary positive coroots are 0 or 1), equivalent to being minimal (and nonzero) in the usual partial ordering of dominant weights. In particular, for an irreducible representation having such a highest weight all its weights are conjugate under the Weyl group, with only the highest weight therefore being dominant.
For irreducible root systems of other than type $A$, the minuscule weights are also fundamental but there are very few of them. For example, the 0 weight always occurs in irreducible representations for type $G_2$ and rules out existence of minuscule weights. Bourbaki is the most standard reference (Groupes et algebres de Lie, VIII, 7.3), but in textbooks the details are usually left as an exercise.
In type A, it's a standard exercise to compute the highest weights occurring in the various exterior powers of the standard representation (whose highest weight is the first fundamental weight) and see directly that these are minuscule and in particular fundamental.
Actually, the semisimplicity should hold with no hypotheses on X, so no example should exist. In fact it is generally expected that, with char. 0 coefficients and over a finite field (both hypotheses being necessary), every mixed motive is a direct sum of pure motives -- so the question for arbitrary varieties reduces to that for smooth projective ones.
The reason is as follows: the different weight-pieces have no frobenius eigenvalues in common (by the Weil conjectures), so the weight filtration can be split by a simple matter of linear algebra. (And the splitting will even be motivic since frobenius is a map of varieties.)
Edit: In response to Jim's comment, let me try to provide a clearer argument (2nd edit: no longer using the Tate conjecture). I claim that if we assume the existence of a motivic t-structure over F_q w.r.t. the l-adic realization in the sense of Beilinson's article http://arxiv.org/pdf/1006.1116v2.pdf, then provided that H^i_c(X-bar) is Frobenius-semisimple for smooth projective X, it is in fact so for aribtrary X.
Indeed, given a motivic t-structure, its heart is an artinian abelian category where every irreducible object is a summand of a Tate-twist of an H^i(X) for X smooth an projective, and furthermore there are no extensions between such irreducibles of the same weight (this is all in Beilinson's article).
That's all true over a general field. But now let's argue that, in the case of a finite field, there also can't be extensions between such irreducibles of different weights; then in the motivic category all of our H^i_c(X-bar) of interest will be direct sums of summands of H^i(X)'s, and we'll have successfully made the reduction to the smooth projective case.
So suppose M and N are irreducible motives of distinct weights over F_q, and say E is an extension of M by N. Consider the characteristic polynomials p_M and p_N of Frobenius acting on the l-adic cohomologies of M and N. By Deligne, they have rational coefficients and distinct eigenvalues, so we can solve q * p_N == 1 (mod p_M) for a rational-coefficient polynomial q. But then (q*p_N)(frobenius) acting on E splits the extension (recall from Beilinson's article that the l-adic realization is faithful under our hypothesis), and we're done.
Later commentary: apparently, when I wrote this I was a little too excited about the perspectives offered by motives. I should emphasize the point essentially made by Minhyong Kim, that the reduction from the general case to the proper smooth case likely doesn't require any motivic technology, and should even be independent of any conjectures. One just needs to know that there's a weight filtration on l-adic cohomology of the standard type where the pure pieces are direct sums of direct summands of appropriate cohomology of smooth projective varieties. As Minhyong says, this probably follows from Deligne's original pure --> mixed argument, via use of compactifications and de Jong alterations. Or at least that's what it seems to me without having gone into the details. I'm sure someone else knows better.
Best Answer
One of the sneaky tricks that Lie theorists play on students is that they tell them about Cartan subalgebras, and then at some point, they pull the rug out and say "just joking; really you should think about the abstract Cartan." The abstract Cartan is a Borel mod its radical. You might say "which Borel?" but it doesn't matter; all these quotients are canonically isomorphic (any two Borels are conjugate, and any two ways of making them conjugate differ by an element of the Borel).
Note, a Cartan subgroup of your Lie algebra isn't canonically isomorphic to the abstract Cartan; you have to choose a Borel containing that Cartan subgroup first. That's where the symmetry is broken.
The abstract Cartan has a canonical notion of positive weight: you look at the action of $B/N$ on the quotient of the Lie algebra $n/[n,n]$, and the weights that appear there are your simple roots (all the positive roots can be obtained by taking the associated graded for the filtration $n\supset [n,n]\supset [n,[n,n]]$). Again, any way of making Borels conjugate carries these weights to the corresponding ones for the other Borel. So once you've picked a Borel, you can use the isomorphism of the abstract Cartan to your chosen one to get a root system on your chosen one.
EDIT: As Allen notes below: I learned most of this from the book of Chriss and Ginzburg.
You get fundamental coroots by taking minimal parabolics over your Borel (those obtained by adding one negative simple root). There's a canonical map from the abstract Cartan of $P/[P,P]$ to that of $G$, and there's unique coweight of $P/[P,P]$ (which is $SL_2$ or $PSL_2$) so that the action on the unique simple root space has eigenvalue 2. The image of that is the simple coroot.