The Kaehler polarisation is a choice of complex structure, $I$, on your torus $M\simeq S^1\times S^1$, which is compatible with the symplectic form. In other words, your torus becomes a complex manifold $X=(M,I)$, and $I$ satisfies $I^t\omega I=\omega$, $g=\omega I>0$. In particular, this tells you that $\omega$ is a positive $(1,1)$-form, i.e., locally
$\omega = \frac{i}{2}h dz \wedge d\overline{z}$, $h>0$. Of course, since $\dim X=1$, $\omega$ is a $(1,1)$-form for any choice of $I$, but this is not so in higher dimensions.
The connection $D$ can now be decomposed as $D= D^{1,0} + D^{0,1}$. Explicitly,
$ D^{0,1} = \frac{1}{2}(1+iI)D$ and $ D^{1,0} = \frac{1}{2}(1-iI)D$. Or, if you prefer, choose a local frame $s$ of $L$, and write
$$
D = d + A_1 dq + A_2 dp = (\partial + B_1 dz) + (\overline{\partial} + B_2 d\overline{z})
$$
Since $\omega$ is of type $(1,1)$, the condition
$D^2 = -i\omega$ translates to
$$
(D^{0,1})^2=0,\quad (D^{1,0})^2=0,\quad D^{0,1}D^{1,0} +D^{1,0} D^{0,1}=-i\omega .
$$
Again, the first two equations hold automatically if $\dim X=1$, but not in general.
This means that on $L$ you get the structure of a holomorphic line bundle $\mathcal{L}$ by taking the Dolbeault operator to be $\overline{\partial}_L= D^{0,1}$. Locally, if you choose a (smooth) frame $s$, this is the above $\overline{\partial}_L =
\overline{\partial} + B_2 d\overline{z}$.
Then the "Hilbert space" is $H^0(X,\mathcal{L})= \ker\overline{\partial}_L \subset A^0(L)$,
where $A^0(L)$ is the (infinite-dimensional) vector space of global smooth sections of $L$. With respect to a local trivialisation, a section $\sigma =fs$ is holomorphic if
$$
\frac{\partial f}{\partial\overline{z}} + B_2 f=0.
$$
The space $H^0(X,\mathcal{L})$ is finite-dimensional for very general reasons. If you actually want to know how big it is, you look at the (Hirzebruch)-Riemann-Roch formula, which tells you that
$$
\dim H^0(X,\mathcal{L})- \dim H^1(X,\mathcal{L}) = \deg \mathcal{L} + 1-g =
\deg \mathcal{L}.
$$
Then $\dim H^0(X,\mathcal{L})=\deg \mathcal{L}$ if $\deg\mathcal{L}>0$, $\dim H^0(X,\mathcal{L})= 0$ if $\deg\mathcal{L}<0$ (or if $\deg\mathcal{L}=0$ and $\mathcal{L}$ is nontrivial), and $\dim H^0(X,\mathcal{L})=1$ if $\mathcal{L}=\mathcal{O}_X$. You can describe the elements of $H^0(X,\mathcal{L})$ quite explicitly
using theta-functions.
Notice that everything - in particular $H^0(X,\mathcal{L})$ - depends on the choice of complex structure $I$, and even the dimension may jump as you vary the complex structure (if $\deg L=0$). But if you choose your prequantum line bundle to be sufficiently positive, the
vector spaces $H^0(X,\mathcal{L})$ "glue" into a vector bundle over the upper half-plane.
ADDENDUM:
Let me spell out the relation with theta-functions a bit more explicitly.
This is a huge classical topic (see the references at the end), so I'll be a bit sketchy.
Write $V=\mathbb{C}$ and $\Lambda = \mathbb{Z}\oplus \tau \mathbb{Z}$, $\textrm{Im}\tau>0$. Then
$\pi: V\to X=V/\Lambda$ is the universal covering map, and it is a local biholomorphism.
If your symplectic structure descends from the form $\omega= c\ dq\wedge dp$ on $V=\mathbb{C}$, the "Bohr-Sommerfeld condition" tells you that $c=\frac{2\pi}{\textrm{Im}\tau}n$,
$n\in \mathbb{Z}$. Notice that $\frac{1}{2\pi}\omega$ is the imaginary part of the hermitian form $H(z,w)=\frac{n}{\textrm{Im}\tau}z\overline{w}$ on $V$ and takes integer values on the lattice.
Suppose $\mathcal{L}$ is as above, with
$c_1(\mathcal{L}) =\frac{1}{2\pi}\left[\omega\right]$, $\deg \mathcal{L}=n$.
As any holomorphic line bundle on $V$ is trivial, we can choose a global holomorphic trivialisation
$\pi^\ast \mathcal{L}\simeq V\times \mathbb{C}$,
and identify $\mathcal{L}= \pi^\ast \mathcal{L}/\Lambda$
with $V\times\mathbb{C}/\sim$. The equivalence relation is
$(z,t)\sim (z+\lambda, a(\lambda,z)t)$, $\lambda\in\Lambda=\pi_1(X)$.
Here $a: \Lambda\times V\to \mathbb{C}^\times$ are the multipliers or factors of automorphy. They are 1-cocyles of $\Lambda=\pi_1(X)$ with values in the entire nonvanishing functions $H^0(V,\mathcal{O}^\times_V)$, i.e., they
are holomorphic in $z$ and
satisfy
$$
a(\lambda+\mu,z)= a(\mu,\lambda+z)a(\lambda,z).
$$
The multipliers do not, in general, take values in $U(1)$!
Changing the trivialisation of $\pi^\ast \mathcal{L}$ replaces $a$ by a coboundary, and
we have
$H^1(\Lambda,H^0(V,\mathcal{O}^\times_V))\simeq H^1(X,\mathcal{O}^\times_X)$.
Now, we can identify $H^0(X,\mathcal{L})\simeq H^0(V,\mathcal{O}_V)^{\pi_1(X)}$.
I.e., we can identify $H^0(X,\mathcal{L})$ with the entire functions which satisfy the
functional equation
$$
\theta(z+\lambda) = a(\lambda,z)\theta(z),
$$
and these are by definition theta-functions.
The explicit structure of the factors of automorphy is known. For instance, in each degree $n$ you have a canonical choice,
$$
a_0(\lambda,z) = \exp\left( n\pi i ab \right)\exp \frac{n\pi}{\textrm{Im}\tau}\left(z\overline{\lambda}+ \frac{1}{2}|\lambda|^2\right),\quad \lambda=a+b\tau.
$$
You can get the other bundles of the same degree by multiplying $a_0$ with characters
$\chi\in \textrm{Hom}(\Lambda,U(1))\simeq Pic^0(X)$.
Finally, a word about the inner product. We can identify $A^0(\mathcal{L})$, the smooth sections of
$\mathcal{L}$, with smooth functions on $V$, satisfying the same functional equation
as the theta-functions. In terms of this identification the fibrewise hermitian product on $\mathcal{L}$ is
$$
\langle f,g \rangle(z) = f(z)\overline{g(z)}\exp \left(-\frac{n\pi}{\textrm{Im}\tau}|z|^2 \right).
$$
The $L^2$-inner product on $H^0(X,\mathcal{L})$ is obtained by integrating this quantity over $X$.
Line bundles on a complex torus are described in great detail in Birkenhake and Lange, Complex Abelian Varieties (especially Chapter 2), Griffiths and Harris, Principles of AG, Chapter 2, Section 6, and
D.Mumford, Abelian Varieties, Chapter 1, Section 2.
Here we have been looking at a 1-dimensional torus, but if you decide to consider the geometric quantisation of $U(1)$ Chern-Simons theory on a higher genus curve, you will need the case of tori of higher dimensions.
Best Answer
The usual answer is that the dual lattice is $\check{\Gamma}=\{f\in V^* | f(\gamma)\in \mathbb{Z}\ \forall \gamma \in \Gamma\}$. It is defined for any lattice $\Gamma\subset V$ - no extra information needed.