$\DeclareMathOperator{im}{im}$

Based on @YangMills suggestion I examined Bruno Bigolin's paper titled "Gruppi di Aeppli" and as a result I am now able to show the following relation

> $$\ker\partial\cap\ker\bar\partial=\im\partial\bar\partial\oplus_\perp\mathcal{H}.$$

Let $\psi\in\ker\partial\cap\ker\bar\partial.$ We can represent $\psi$ as 
$$\psi=\Delta G\psi+\pi\psi=\\\color{red}{\partial\bar\partial\bar\partial^*\partial^* G\psi}+\color{blue}{\bar\partial^*\partial^*\partial\bar\partial G\psi}+\color{blue}{\bar\partial^*\partial\partial^*\bar\partial G\psi}+\color{green}{\partial^*\bar\partial\bar\partial^*\partial G\psi}+\color{blue}{\bar\partial^*\bar\partial G\psi}+\color{green}{\partial^*\partial G\psi} + \pi\psi=\\\color{red}{\partial\bar\partial\alpha}+\color{green}{\partial^*\lambda}+\color{blue}{\bar\partial^*\mu}+h,$$
where $G$ is Green's operator and $\pi$ is projection on $\mathcal{H}.$
As a result $$\partial^*\lambda+\bar\partial^*\mu=\psi-\partial\bar\partial\alpha-h.$$ Now since $\psi,\partial\bar\partial\alpha,h\in\ker\partial\cap\ker\bar\partial$, so as $\partial^*\lambda+\bar\partial^*\mu\in\ker\partial\cap\ker\bar\partial.$ And then
$$||\partial^*\lambda+\bar\partial^*\mu||^2=\langle\partial(\partial^*\lambda+\bar\partial^*\mu),\lambda\rangle+\langle\bar\partial^*(\partial^*\lambda+\bar\partial^*\mu),\mu\rangle=0.$$
Hence
$$\psi=\partial\bar\partial\alpha+h.$$
And yet $\im\partial\bar\partial\perp\mathcal{H}.$ In fact
$$\langle\partial\bar\partial\alpha,h\rangle=\langle\alpha,\bar\partial^*\partial^*h\rangle=0.$$
So we proved that
$$\ker\partial\cap\ker\bar\partial=\im\partial\bar\partial\oplus_\perp\mathcal{H}.$$
Immediate consequence of above is that
$$H_{BC}\cong\mathcal{H}.$$
And since $\Delta$ is elliptic we get that $\dim_\mathbb{C}H_{BC}<\infty.$

PS. Above $\Delta=\tilde\Delta_{BC}^{p,q},\mathcal{H}=\mathcal{H}^{p,q}_{\tilde\Delta_{BC}}, H_{BC}=H^{p,q}_{BC}$ and $\psi\in\mathcal{E}^{p,q}.$