There is strong circumstantial evidence that the question came from a personal conversation with Bogomolov. First, Gromov's paper has more than 50 references, but he doesn't cite Bogomolov at all. Second, he does later in the paper thank Bogomolov for a helpful conversation, referring to him in nickname form as Fedya. They are about the same age, they both have appointments at NYU, and Gromov has thanked Bogomolov in papers before. Third, when I skimmed over Bogomolov's papers (okay, the abstracts and some introductions), I couldn't find anything directly bearing on the question.
On the other hand, I did find a paper by Carlson and Toledo that said that 30 years ago, Gromov introduced the partial ordering on $n$-manifolds given by the existence of a map $M \to N$ with non-zero degree. The broad question is what this partial ordering looks like. Obviously Gromov norm is an important invariant for this question.
Unless you ask Gromov or Bogomolov directly, it may make more sense to make an educated guess based on Gromov's summary. I would say, if $X$ is a closed, smooth, $n$-dimensional algebraic variety over $\mathbb{C}$, is there a closed, smooth $n$-fold $Y$ with a non-constant morphism $Y \to X$, such that $Y$ also fibers smoothly over a complex curve? Unless there is some clear reason that this question is misstated, it surely works as motivation for Gromov's question.
Let $X$ be a simplicial set. The partially ordered set $(X^{nd}, \leq)$ of non-degenerate simplices of $X$, with $x\le y$ if $x$ is a face of $y$, was considered by Barratt in a 1956 Princeton preprint. I write $B(X) = N(X^{nd}, \le)$ for its nerve, the Barratt nerve.
It is not quite clear to me if, in your definition of $Face \colon sSet \to Pos$, you only allow the non-degenerate simplices as elements. Let me assume that you make that restriction, so that $Nerve \circ Face(X) = B(X)$.
If $X$ is the simplicial set associated to an (ordered) simplicial complex, then $B(X)$ is
indeed the barycentric subdivision of that simplicial complex. In these cases, there is a homeomorphism $|B(X)| \cong |X|$. However, if $X = \Delta^n/\partial\Delta^n$ for $n\ge1$,
then $X^{nd}$ is isomorphic to $[1] = (0 < 1)$ and its nerve is contractible. So in these cases $|B(X)|$ is not homeomorphic (nor homotopy equivalent) to $|X|$.
From the definition of Kan's normal subdivision $Sd(X)$ as a coend, there is a canonical
natural map $b_X \colon Sd(X) \to B(X)$. It is an isomorphism if and only if $X$ is non-singular, meaning that the representing map $\bar x \colon \Delta^n \to X$ of each non-degenerate simplex of $X$ is a cofibration. Here $n$ is the dimension of $x$.
There exists a homeomorphism $h_X \colon |Sd(X)| \cong |X|$, due to Fritsch and Puppe (1967). However, it is not natural for most maps $X \to Y$ of simplicial sets. For instance, there is no way to fix $h_{\Delta^1}$ and $h_{\Delta^2}$ so that the homeomorphisms are compatible with both of the degeneracy maps $\sigma^j \colon \Delta^2 \to \Delta^1$.
The normal subdivision $Sd(X)$ of any simplicial set is a regular simplicial set, in the sense that each representing map $\bar x \colon \Delta^n \to X$ only makes identifications
along the last ($n$th) face. The geometric realization of a regular simplicial set
is a regular CW complex, i.e., if the closure of each $n$-cell is an $n$-ball and the boundary of the open $n$-cell in its closure is an $(n-1)$-sphere. Any regular CW complex admits a triangulation. See e.g. Fritsch and Piccinini (1990), sections 3.4 and 4.6 for proofs.
Combining the two previous paragraphs, the realization of any simplicial set can be
triangulated by a simplicial complex, but not in a natural way. Not every CW complex can be triangulated (Metzler, 1967), but the realizations of (regular) simplicial sets can.
In work of Waldhausen, Jahren and myself (Spaces of PL Manifolds and Categories of Simple Maps,
Annals of Maths. Studies, to appear in 2013) we consider finite simplicial sets $X$
and prove in Proposition 2.5.8 that there are natural maps
$$
Sd(Sd(X)) \to B(Sd(X)) \to Sd(X) \to X .
$$
Each of these are simple, in the sense that their geometric realizations have contractible
point inverses. In particular, they are simple-homotopy equivalences. The term
$B(Sd(X))$
is the nerve of a partially ordered set, hence an ordered simplicial complex. So there
is, indeed, a natural simple map
$$
i_X \colon B(Sd(X)) \to X
$$
from an ordered simplicial complex to $X$.
You mentioned geometric topology at the outset. If $X$ is a combinatorial manifold,
so that $M = |B(Sd(X))|$ and $N = |X|$ are PL manifolds, then Cohen (1970) proved that
any simple PL map $M \to N$ can be uniformly approximated by a PL homeomorphism (assuming
the manifolds are of dimension at least 5). Hence the natural map $|i_X| \colon |B(Sd(X))| \to |X|$ is a map from an ordered simplicial complex that is arbitrarily
close to a PL homeomorphism, but I have no reason to expect that this
approximation can be made natural.
Best Answer
In brief, the answer to this part of the question is 'You can take $M\to\Sigma_g$ to be regular, but beyond that, no.' This can already be extracted from Misha's comments above, but let me try to summarise.
First, note that there is a regular covering $M\to\Sigma_g$ that factors through $\Sigma_h\to\Sigma_g$. (Specifically, you can take $\pi_1M$ to be the intersection of all the conjugates of $\pi_1\Sigma_h$.) So, as in the earlier part of your question, you can take $\Sigma_h\to\Sigma_g$ to be regular.
You can now rephrase your question in terms of normal quotients of $\pi_1\Sigma_g$, and it becomes
In particular, $P$ surjects $Q$. But any finite group can arise as $Q$ (see Misha's and algori's comments---the point is that $\pi_1\Sigma_g$ surjects a free group), so you are looking for a 'standard' family of finite groups that surjects every finite group. But there's no 'natural' definition of such a family.
Remark: Obviously, there are such families, such as $\{ Q\times\mathbb{Z}/2\}$ where $Q$ is an arbitrary finite group, but clearly this is not 'natural'.