First a small thing. I am pretty sure we don't have $K_2(\mathbb C((t)))=\mathbb
C^*$, we have a surjective residue homomorphism $K_2(\mathbb C((t)))\rightarrow
\mathbb C^*$ but, I believe, with a non-trivial kernel. In any case, we can look
at the induced central extension and then the rest of what you say is
OK. Similarly, we have a surjective map $K_2(\mathbb Q_p)$.

Disrergarding this, there is a much simpler analogy between the two cases which on the
one hand, I think, makes the analogy that you want less likely and on the other
hand can be proven... To begin with it is not quite true that even $G(\mathbb
C((t)))$ is defined over $\mathbb C$ at least not as a group scheme. What
happens is that $G(\mathbb C[[t]])$ is a group scheme, it is the inverse limit
of the $G(\mathbb C[t]/(t^n))$ and these have a natural structure of algebraic
group over $\mathbb C$ (through the Greenberg functor). then $G(\mathbb C[[t]])$
as the inverse limit of algebraic groups is a group scheme (it is not of finite
type hence convention forces us to call it a group scheme rather than algebraic
group). Now, if we try to pass to $G(\mathbb
C((t)))$ we get into trouble. It is an infinite union of schemes (bound the
valuations of the entries of the elements of $G(\mathbb
C((t)))$ in some faithful linear representation of $G$) but an infinite union of
schemes does in general not have a scheme structure. There are ways of extending
the scheme notion to cover this case and what we get is what is called an
ind-group scheme over $\mathbb C$. Also the loop group type extension of $G(\mathbb
C((t)))$ by $\mathbb C^*$ has such an extension (as does every Kac-Moody type
group).

The situation for $G(\mathbb Q_p)$ is almost identical; $G(\mathbb Z/p^n)$ are
the $\mathbb Z/p$-points of a $\mathbb Z/p$-algebraic group, $G(\mathbb Z_p)$
are the $\mathbb Z/p$-points of a group scheme over $\mathbb Z/p$ and $G(\mathbb
Q_p)$ are the $\mathbb Z/p$-points of an ind-group scheme over $\mathbb Z/p$. I
think that the same thing is true for the central extension. The upshot is that
there is a close analogy to the $\mathbb C$ case but in that analogy $\mathbb C$
is replaced by $\mathbb F_p$ not by $\mathbb F_1$.

Note that in the Connes-Consani version of $\mathbb F_1$ $G$ is defined over $\mathbb F_{1^2}$ so perhaps that is the place to look for a version of the Brylinski-Deligne result.

<b>Addendum</b>: Just to add even more concreteness to George's answer about the explicit form of Greenberg's functor for $\mathbb G_m$. We have that $W_n(B)$ is just $B^n$ with a funny multiplication and addition. They are however given by polynomials (which are independent of $B$). The units in this ring are the tuples of the form $B^\ast\times B^{n-1}$ and multiplication is given by polynomials. This means that the algebraic group associated to this is just $\mathbb G_m\times\mathbb A^{n-1}$ as scheme but with a funny product structure. In particular its $\mathbb F_p$-points are just $(\mathbb Z/p^n)^\ast$.