This remark is just a caution for when one tries to think about what the Spec of $U\mathfrak g$ might mean.
Suppose that $\mathfrak g$ is semisimple, and let $I$ be the kernel of the natural augmentation $U\mathfrak g \to \mathbb C$. (Alternatively, $I$ is the kernel of the action of $U\mathfrak g$ on the trivial representation of $\mathfrak g$.) From the semisimplicity one finds that $I^2 = I$, and hence that $I^n = I$ for every $n \geq 1$.
One also checks (e.g. by a grading argument) that $U\mathfrak g$ is a non-commutative domain, in the sense that it has no zero divisors, and that $U\mathfrak g$ is Noetherian.
Now one easily checks that if $A$ is a commutative Noetherian domain, and if $I$ is
a non-unit ideal such that $I^2 = I$, then necessarily $I = 0$. The point is that if
$I = I^2$, the zero-locus of $I$ in Spec $A$ is "formally" isolated from its complement,
and so morally provides a disconnection of Spec $A$ (and this moral argument is made rigorous via an application of the Krull intersection theorem). If $A$ is a domain, on the other hand, then Spec $A$ is irreducible, and so in particular connected, and so the only possibility
is that $I = 0$.
So $U\mathfrak g$ has various properties that would be mutually incompatible in a commutative ring, and hence one has to be careful in importing geometric intuition naively
in thinking about some kind of non-commutative Spec $U\mathfrak g$.
(Of course, the other answers here give a very non-naive discussion of various geometric points of view on $U\mathfrak g$, but I hope that the above remark will be useful to some people. Let me note that it is also somewhat related to this answer.)