The following ought to answer your question.

**Prop.** Let $\mathcal{X}$ be a smooth proper complex analytic space. We have a long exact sequence of abelian groups:

$$\cdots \to H^i_{\mathcal{D}}(\mathcal{X},\mathbf{Z}(j))\to H^i(\mathcal{X},\mathbf{Z})\to H^i(\mathcal{X},\mathbf{C})/F^jH^i(\mathcal{X},\mathbf{C})\to H^{i+1}_{\mathcal{D}}(\mathcal{X},\mathbf{Z}(j))\to\cdots$$

where $F^{\bullet}H^i(\mathcal{X},\mathbf{C})$ is the Hodge filtration on $H^i(\mathcal{X},\mathbf{C})$.

**Sketch of Proof.**
By design of the Deligne complex $\mathbf{Z}(j)_{\mathcal{D}}$, we have a triangle in $D(\text{Ab})$:
$$\Omega_{\mathcal{X}/\mathbf{C}}^{\le j-1}[1]\to \mathbf{Z}(j)_{\mathcal{D}}\to\mathbf{Z}.$$
Form hypercohomology and prove $\mathbb{H}^i(\mathcal{X},\Omega_{\mathcal{X}/\mathbf{C}}^{\le j-1}) = H^i(\mathcal{X},\mathbf{C})/F^jH^i(\mathcal{X},\mathbf{C})$.
To see this, use the triangle:
$$\Omega_{\mathcal{X}/\mathbf{C}}^{\ge j}\to \Omega_{\mathcal{X}/\mathbf{C}}^{\bullet}\to \Omega_{\mathcal{X}/\mathbf{C}}^{\le j-1}$$
and reduce to check $\mathbb{H}^i(\mathcal{X},\Omega_{\mathcal{X}/\mathbf{C}}^{\ge j})=F^jH^i(\mathcal{X},\mathbf{C})$, which follows almost by definition of the Hodge filtration. QED

In particular, for $i=2j$, we get:
$$0\to J^j(\mathcal{X})\to H^{2j}_{\mathcal{D}}(\mathcal{X},\mathbf{Z}(j))\to \text{Hdg}^{j}(\mathcal{X},\mathbf{Z})\to 0$$
upon being careful about grading conventions (eg. I have seen $J^n$ often denoted $J^{2n}$, etc.)

**Rem.** Note that $H^i(\mathcal{X},\mathbf{Z})$ is always finitely generated, hence your question really is about when the image of $H^i(\mathcal{X},\mathbf{Z})$ in $H^i(\mathcal{X},\mathbf{C})/F^jH^i(\mathcal{X},\mathbf{C})$ under the above edge map, is a full lattice. If $i$ is odd this is true.