According to @Donu Arapura's comment, I give an answer of my understanding, whether it is correct or not, please comment below.

By the definition of Frölicher spectral sequence degenerating at $E_1$, we have particularly $E_1^{0,2}=E_{\infty}^{0,2}$, which is equivalent to $H^{0,2}_{\bar\partial}(X)=\frac{H^2(X,\mathbb C)}{F^1H^2(X,\mathbb C)}$, where $F^1H^2(X,\mathbb C):=\frac{F^1A^2(X)\cap\ker d}{F^1A^2(X)\cap\text{im }d}$. Then the map $H^2(X,\mathbb C)\to H^2(X,\mathcal O)$ becomes $H^2(X,\mathbb C)\to \frac{H^2(X,\mathbb C)}{F^1H^2(X,\mathbb C)}$, which seems obviously surjective.

According to @Donu Arapura's comment, I give an answer of my understanding, whether it is correct or not, please comment below.

By the definition of Frölicher spectral sequence degenerating at $E_1$, we have particularly $E_1^{0,2}=E_{\infty}^{0,2}$, which is equivalent to $H^{0,2}_{\bar\partial}(X)=\frac{H^2(X,\mathbb C)}{F^1H^2(X,\mathbb C)}$, where $F^1H^2(X,\mathbb C):=\frac{F^1A^2(X)\cap\ker d}{F^1A^2(X)\cap\text{im }d}$. Then the map $H^2(X,\mathbb C)\to H^2(X,\mathcal O)$ becomes $H^2(X,\mathbb C)\to \frac{H^2(X,\mathbb C)}{F^1H^2(X,\mathbb C)}$, which seems obviously surjective.