Using the Kummer exact sequence
$0\rightarrow\mu_n\rightarrow\mathbb{G}_m\rightarrow\mathbb{G}_m\rightarrow0$ we
get a long exact sequence
$$
0\rightarrow\mathrm{Hom}(\mathbb{G}_m,A)\rightarrow\mathrm{Hom}(\mathbb{G}_m,A)
\rightarrow\mathrm{Hom}(\mu_n,A)\rightarrow\mathrm{Ext}^1(\mathbb{G}_m,A).
$$
As $\mathrm{Hom}(\mathbb{G}_m,A)=0$ this gives an embedding
$\mathrm{Hom}(\mu_n,A)\hookrightarrow\mathrm{Ext}^1(\mathbb{G}_m,A)$ and at
least over an algebraically closed field in which $n$ is invertible we always
have that $\mathrm{Hom}(\mu_n,A)$ is non-trivial and consequently  so is
$\mathrm{Ext}^1(\mathbb{G}_m,A)$.

This does not contradict Deligne's claim as he seems to be (somewhat implicitly
it must be admitted) speaking of extensions up to isogeny (in Principle 2.1 this
is made more clear). In fact Chevalley's theorem implies that this is true:
Consider a connected affine subgroup $G'\hookrightarrow G$ for which the
quotient is an abelian variety. Then the kernel of the composite
$G'\rightarrow\mathbb{G}_m$ is affine and embeds in $A$ and thus is finite. On
the other hand $G'\rightarrow\mathbb{G}_m$ must be surjective as otherwise $G'$
would be finite and hence trivial so $G$ would be an abelian variety which is
not possible. The kernel of $G'=\mathbb{G}_m\rightarrow\mathbb{G}_m$ is then
some $\mu_n$ which shows that the extension is in the image of
$\mathrm{Hom}(\mu_n,A)\hookrightarrow\mathrm{Ext}^1(\mathbb{G}_m,A)$ and in
particular is trivial up to isogeny.

The final conclusion is that $\mathrm{Ext}^1(\mathbb{G}_m,A)=\mathrm{Hom}(\hat{\mathbb Z}(1),A)$.