Is Deligne cohomology the motivic cohomology of analytic spaces? Let $X$ be a smooth projective complex analytic space.
We can cook up a complex analytic version of Bloch's cycle complex by declaring
$z^n(X^{\rm an}, m)$
is the free abelian group on all codimension $m$ analytic cycles on $X\times\Delta^n$ ($\Delta^n$ being the usual standard $n$ simplex in complex analytic spaces, ie. the spectrum of $\mathbf{C}\{u_0,\ldots, u_n\}/(u_0+\ldots+u_n -1)$) in good position (ie. intersecting every face in the appropriate codimension, as in Bloch's paper). The differential $d_m$ is the same as in Bloch's original definition, turning $(z^n(X^{\rm an}, m), d_m)$ into a complex of abelian groups.
Call $$\mathbf{Z}(n)_{\mathcal{M}} := (z^n(X^{\rm an}, m), d_m)[2m]$$
and its hypercohomology "motivic cohomology of $X$".
Here's the question. Is motivic cohomology of $X$ at all related to the Deligne cohomology of $X$? More optimistically, does there exist a quasi-isomorphism
$$\mathbf{Z}(n)_{\mathcal{M}}\to\mathbf{Z}(n)_{\mathcal{D}} ?$$
How should one think about Deligne cohomology, in other words? (if not as "the motivic cohomology of complex analytic spaces?)
Remarks
I can imagine a regulator map $\text{reg} : \mathbf{Z}(n)_{\mathcal{M}}\to\mathbf{Z}(n)_{\mathcal{D}}$ can be defined using currents, as done for the classical regulator.
This is for sure going to be a (rather uninteresting) quasi isomorphism, since $X$ is smooth, when $n = 0$.
For $n = 1$ this is likely going to be a quasi-isomorphism too (if one doesn't screw the definition of $\text{reg}$): both sides are just $\mathbf{G}_m[-1]$.
 A: The answer is no. 
So far, you have checked only very special cases, such as weight $n$ and degree $2n$, for $n = 0,1$.
Consider the analytic hypercohomology of your $\mathbf{Z}(n)_{\mathcal{M}}$ in degree $2n$, denoted $H^{2n}_{\rm an}(X,\mathbf{Z}(n))$. The same (elementary) argument as in several references (eg. Bloch's "Algebraic Cycles and Higher $K$-Theory", 1986) shows
$$H^{2n}_{\rm an}(X,\mathbf{Z}(n)) \simeq\text{CH}^n(X^{\rm an})$$
the Chow group of analytic cycles on $X^{\rm an}$. By GAGA, there is a canonical isomorphism of abelian groups $\text{CH}^n(X)\simeq\text{CH}^n(X^{\rm an})$.
Upon running the same construction of the cycle map to Deligne cohomology as in Kerr-Lewis, as you suggest, you must end up with a cycle map
$$\text{CH}^n(X^{\rm an})\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$$
that, under the above isomorphism from GAGA, should better agree with the usual cycle map $\text{CH}^n(X)\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$. 
Rem. If your conjecture were true, then, in particular, the cycle map $\text{CH}^n(X)\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$ should be surjective, this latter group being surjecting onto the finitely generated abelian group of Hodge cycles $\text{Hdg}^{n,n}(X^{\rm an})$, the composition being the cycle map to Betti cohomology. As a consequence, the Hodge conjecture is implied by your contention.
There are two problems. I am not aware of any expectation along the lines of surjectivity of the cycle map onto Deligne(-Beilinson) cohomology. Rather, the only claim being usually made is surjectivity onto the abelian group of Hodge classes.
On the other hand, if your contention were correct, then the cycle map $\text{CH}^n(X)\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$ should be injective too, which rarely happens, as this puts strong constraints on the Hodge numbers of $X^{\rm an}$. 
A paper of Esnault and Levine proves that if such cycle map is injective, then it is also surjective, and implies strong conditions on the shape of the Hodge diamond of $X^{\rm an}$. There are many examples of an $X$ that does not meet such conditions.
A: Wouldn't your motivic cohomology group in degree $(2m, m)$ be the Chow group of codimension $m$ cycles? But the corresponding Deligne cohomology group surjects onto the Hodge classes in cohomological degree $2m$. For them to be  quasi-isomorphic you would (at least) need the Hodge conjecture, which is false for a general Kahler manifold.   
As you said any explicit formula for the usual regulator should extend to this setting.  The paper of  Kerr, Lewis, Mueller-Stach https://arxiv.org/abs/math/0409116 does this at least for nullhomologous  classes.
