Basis-free definition of Casimir element?

The Casimir element is dual to the Killing form. (I think. I am somewhat uncertain about this because nobody has ever said this to me, even though it seems like the right thing to say, and frankly I don't know why Lie algebra textbooks don't just say this.) That is, the nondegeneracy of the Killing form is equivalent to its providing an isomorphism $\mathfrak{g} \to \mathfrak{g}^{\ast}$, and writing this isomorphism as a tensor exhibits it as an element of $\mathfrak{g} \otimes \mathfrak{g}$ - precisely the element $\sum e_i \otimes e_i'$ where $e_i$ is a basis. This embeds into $U(\mathfrak{g}) \otimes U(\mathfrak{g})$ and the multiplication map into $U(\mathfrak{g})$ gives the Casimir element.

(Oh good, I see that this is the same definition Akhil gives in the blog post darij linked to above. That makes me feel better.)

Also, you're using the "wrong" definition of the determinant. The exterior powers are all functors, and they take linear maps $T : V \to V$ to linear maps $T : \Lambda^n V \to \Lambda^n V$. Since $\Lambda^n V$ is one-dimensional when $n = \dim V$, the linear maps from $\Lambda^n V$ to itself are canonically isomorphic to the base field $k$. There is also a slightly more transparent definition of the trace: $\text{End}(V)$ is canonically isomorphic to $V^{\ast} \otimes V$, and then one composes with the dual pairing $V^{\ast} \times V \to k$.

It seems to me the notion you're looking for is what general notion of monoidal category supports the definition you're looking at. Traces can be defined in monoidal categories with duals, although I'm not sure what the natural setting for determinants is.

This is an answer to the question about the general structure of the centre of the enveloping algebra:

The centre of the enveloping algebra (in particular when $\mathfrak g$ is reductive) is one of the basic objects in the study of infinite-dimensional representations of Lie algebras and Lie groups. It was first described in general (for reductive $\mathfrak g$) by Harish-Chandra, as far as I know, who proved the following: let $\mathfrak g$ be a reductive Lie algebra (over $\mathbb C$, say) and let $\mathfrak h$ be a Cartan subalgebra. Let $W$ be the Weyl group, which acts on $\mathfrak h$ via the adjoint action.

Then there is an isomorphism $Z(\mathfrak g) \cong Sym(\mathfrak h)^W.$ (Here $Z(\mathfrak g)$ denotes the centre of the enveloping algebra $U(\mathfrak g)$, and $Sym(\mathfrak h)$ denotes the symmetric algebra of $\mathfrak h,$ which is the same as the enveloping algebra $U(\mathfrak h)$, since $\mathfrak h$ is abelian.)

E.g. if $\mathfrak g = {\mathfrak sl}_2$, then a Cartan subalgebra is one-dimensional, therefore $Sym(\mathfrak h)^W$ turns out to be a polynomial ring in one generator, where this generator can be taken to be the Casimir.

If $\mathfrak g$ is semi-simple of rank $l$, then the centre turns out to be a polynomial ring in $l$ generators, one of which is the Casimir, and the others of which can be taken to be the so-called higher Casimirs. (Their degrees are the so-called exponents of $\mathfrak g$, or perhaps the exponents shifted by one; I'm unsure about the standard normalization. [Added: As Mike Skirvin confirms in a comment below, they are the exponents shifted by $1$.])

To learn more you can google Harish-Chandra isomorphism, or look in one of the many representation-theory texts that are out there. (I like Knapp's book Representation Theory of Semisimple Groups: An Overview Based on Examples, but there are lots of choices.)

Concerning your first question: it's really an issue about finite dimensional vector spaces.

(1). If $V$ is a finite dimensional vector space over a field $k$, then one has an evaluation map

$$ e: V^\ast \otimes V \to k $$

defined by $f\otimes v \mapsto f(v)$. Then $$ e \in (V^{\ast} \otimes V)^\ast . $$ There is a canonical isomorphism $V \otimes V^\ast\to (V^\ast \otimes V)^\ast$ given by $$ (v,f) \mapsto ((g,w) \mapsto g(v)f(w)) . $$ Using this isomorphism, we can regard $e$ corresponds to element $e'\in V\otimes V^\ast$. This is the Casimir element.

Remark: $e'$ is characterized by the following property: it is the unique vector in $V\otimes V^{\ast}$ such that for all vector spaces $W$, the map $$ \hom(V,W) \to W\otimes V^\ast $$ given by $f\mapsto (f\otimes 1_{V^\ast})(e')$ is an isomorphism.

(2). More generally, if one is given a perfect pairing: $$ e : U \otimes V \to k $$ (meaning that the adjoint $U \to V^\ast$ is an isomorphism), one gets by the same procedure an element $e'\in V\otimes U$.

(3). $V = \frak g$ is a semi-simple Lie algebra, then the killing form $V\otimes V \to k$ is perfect. So we get by (2) the (Casimir) element $e' \in V\otimes V$.