I'll show where the sequence comes from over $\mathbb{C}$ and then point out why some of the obvious ways people try to generalize to other settings don't yield anything immediately.
The sequence expressed above comes from the exponential sequence $0\to 2\pi i\mathbb{Z}\to \mathcal{O}_X\stackrel{exp}{\to} \mathcal{O}_X^*\to 0$. We take the long exact sequence associated to this and get
$H^1(X, \mathbb{Z})\hookrightarrow H^1(X, \mathcal{O}_X)\to H^1(X, \mathcal{O}_X^*)\stackrel{c_1}{\to} H^2(X, \mathbb{Z})\to $
Now $H^1(X,\mathcal{O}_X^*)=Pic(X)$ and $c_1$ is the first Chern class which in cases where it makes sense corresponds to the degree of the line bundle. Thus taking $Pic^0(X)$ to be line bundles of Chern class $0$, we get the sequence you wanted $0\to H^1(X, \mathbb{Z})\to H^{0,1}(X)\to Pic^0(X)\to 0$.
Suppose $X/k$ is smooth and projective. We have lots of issues in generalizing right off. What do we want $Pic^0(X)$ to mean? How should we replace cohomology with $\mathbb{Z}$ coefficients? What do we do about the exponential sequence which no longer exists?
The standard fix for the exponential sequence is to use the Kummer sequence in etale cohomology. Suppose $char(k)=p$ and choose a prime $\ell\neq p$. We have an exact sequence $0\to \mathbb{Z}/\ell^n \to \mathbb{G}_m\stackrel{\ell^n}{\to} \mathbb{G}_m\to 0$
The long exact sequence in etale cohomology gets us the "right" Betti numbers and "integral" coefficients, but fails to produce something that looks like the sequence you want. We get by taking the inverse limit over all $n$:
$\to H^1_{et}(X_{\overline{k}}, \mathbb{Z}_\ell(1))\to Pic(X)\to Pic(X)\stackrel{c_1}{\to} H^2_{et}(X_{\overline{k}}, \mathbb{Z}_\ell(1))\to $
Where $c_1$ is now the first etale Chern class. We could do a similar thing and make $Pic^0(X)$ the line bundles with Chern class $0$ to get surjective on the end. As you see we won't get any sort of Hodge decomposition though.
Another attempted fix would be to assume the Hodge-de Rham spectral sequence degenerates and hence $H_{dR}(X/k)\simeq H^{1,0}\bigoplus H^{0,1}$ to get some sort of Hodge part to the sequence, but I don't see how to get it into a sequence without an exponential map because the standard projection is clearly going to have kernel $H^{1,0}$ and not be injective.
This doesn't answer your question, but I thought writing it out might help someone else figure it out.