Here is how I like to understand the Weil pairing.

The dual of an abelian variety $A$ is the scheme $\hat{A} = \mathrm{Hom}(A, B\mathbf{G}_m)$. Here $B\mathbf{G}_m$ is the stack of line bundles and $\mathrm{Hom}$ refers to homomorphisms of group stacks. One therefore has a perfect pairing

$A \times \hat{A} \rightarrow B\mathbf{G}_m$.

This actually means something relatively concrete: For each pair of points $(a, a') \in A \times \hat{A}$ one has a one dimensional vector space $L(a,a')$, together with isomorphisms

$L(a_1 + a_2, a') \simeq L(a_1, a') \otimes L(a_2, a')$,

$L(a, a'_1 + a'_2) \simeq L(a, a'_1) \otimes L(a, a'_2)$

satisfying some compatibility conditions (you can find the details of this definition under the heading of biextensions; there is a nice explanation in SGA7). One of these compatibilities is that the two isomorphisms

$L(a_1 + a_2, a'_1 + a'_2) \simeq L(a_1, a'_1) \otimes L(a_1, a'_2) \otimes L(a_2, a'_1) \otimes L(a_2, a'_2)$

should coincide.

One consequence of the definition is that $L(0, a') = L(0 + 0, a') \simeq L(0, a') \otimes L(0, a')$ which means that $L(0, a')$ is canonically trivialized, as is $L(a, 0)$ by symmetry. Moreover, the two trivializations of $L(0,0)$ must be the same.

If we choose an $n$-torsion point $a$ of $A$ then $L(a, a')$ will be a line bundle with a trivialization of its $n$-th tensor power. Similarly, if $a'$ is also an $n$-torsion point then $L(a,a')^{\otimes n}$ will come with *two* trivializations coming from its identification with the canonically trivialized line bundles $L(na, a') = L(0,a')$ and $L(a, na') = L(a,0)$. Comparing these two trivializations, we get an element of $\mathbf{G}_m$ and therefore a pairing

$A[n] \times \hat{A}[n] \rightarrow \mathbf{G}_m$.

However, we notice that the induced trivializations of $L^{\otimes n^2} \simeq L(na, na')$ must coincide. Therefore the image of this map actually lands in $\mathbf{G}_m[n] = \mu_n$. This gives the Weil pairing

$A[n] \times \hat{A}[n] \rightarrow \mu_n$.

I don't know of a way to interpret the Hilbert symbol in quite this way, but I'd be very interested if someone could suggest one!