By the way, when you plug in $\mathcal{V}= Set$, then you essentially get the Simpson/Tamsamani version of weak n-category. The comparison of this with bicategories is describe several places, for example in Leinster's survey. Let's spell it out a bit. We begin with 0-categories which are just sets. An equivalence of sets is a bijection. This means that the Segal maps become isomorphisms. It is a standard exercise that this reproduces the usual notion of category.

Next, on the second iteration, a weak 2-category has a set of object and then for each pair a hom category. There are also categories $C(x,y,z)$, etc. and $C(x,y,z) \to C(x,y) \times C(y,z)$ is an equivalence. To get a composition functor we must make choices of inverse equivalences to these. By looking at the diagram we get for quadruples of objects, we get an associator, and by looking at the diagram for quintuples, we see that the associator satisfies the pentagon equation.

So you can get a bicategory, but the correspondence is not quite 1-1. This is discussed in more detail in this paper by Lack and Paoli.

By the way, when you plug in $\mathcal{V}= Set$, then you essentially get the Simpson/Tamsamani version of weak n-category. The comparison of this with bicategories is described in several places, for example in Leinster's survey. Let's spell it out a bit. We begin with 0-categories which are just sets. An equivalence of sets is a bijection. This means that the Segal maps become isomorphisms. It is a standard exercise that this reproduces the usual notion of category.

Next, on the second iteration, a weak 2-category has a set of object and then for each pair a hom category. There are also categories $C(x,y,z)$, etc. and $C(x,y,z) \to C(x,y) \times C(y,z)$ is an equivalence. To get a bicategory from this we need a composition functor, and to get that we must make choices of inverse equivalences to these maps. By making similar choices for the diagram we get for quadruples of objects, we get an associator, and by looking at the diagram for quintuples, we see that the associator satisfies the pentagon equation.

So you can get a bicategory, but the correspondence is not quite 1-1. However it is up to equivalence (i.e., the bicategories resulting from different choices will be equivalent as bicategories). This is discussed in more detail in this paper by Lack and Paoli.

First, as Rune pointed out in the comments, his paper with David Gepner gives a very general approach to your wish list. However to make it so general that it applies to arbitrary monoidal $(\infty,1)$-categories (whatever those are...) means that the machinery is arguably very complicated. In particularly it probably doesn't satisfy your (4) (it is not "effortless").

Plus there are maps which are like the maps of a simplicial set, and some deriviative maps from these are required to be equivalences (the "Segal maps").

By the way, when you plug in $\mathcal{V}= Set$, then you essentially get the Simpson/Tamsamani version of weak n-category. The comparison of this with bicategories is describe several places, for example in Leinster's survey.

First, as Rune pointed out in the comments, his paper with David Gepner gives a very general approach to your wish list. However to make it so general that it applies to arbitrary monoidal $(\infty,1)$-categories (whatever those are, ) means that the machinery is arguably very complicated. In particular, it probably doesn't satisfy your (4). It is not "effortless".

Plus there are maps which are like the maps of a simplicial set. So for example there are degeneracy/unit maps like $1 \to C(x,x)$ and $C(x,y) \to C(x,x,y)$ which add identities, and there are face/restriction maps like $C(w,x,y,z) \to C(x,y,z)$ or $C(w,x,y,z) \to C(w,y,z)$.

Some derivative maps from these are required to be equivalences (the "Segal maps"). Specifically the maps $$ C(w,x,y,z) \to C(w,x) \times C(x,y) \times C(y,z) $$ and its cousins.

By the way, when you plug in $\mathcal{V}= Set$, then you essentially get the Simpson/Tamsamani version of weak n-category. The comparison of this with bicategories is describe several places, for example in Leinster's survey. Let's spell it out a bit. We begin with 0-categories which are just sets. An equivalence of sets is a bijection. This means that the Segal maps become isomorphisms. It is a standard exercise that this reproduces the usual notion of category.

Next, on the second iteration, a weak 2-category has a set of object and then for each pair a hom category. There are also categories $C(x,y,z)$, etc. and $C(x,y,z) \to C(x,y) \times C(y,z)$ is an equivalence. To get a composition functor we must make choices of inverse equivalences to these. By looking at the diagram we get for quadruples of objects, we get an associator, and by looking at the diagram for quintuples, we see that the associator satisfies the pentagon equation.

So you can get a bicategory, but the correspondence is not quite 1-1. This is discussed in more detail in this paper by Lack and Paoli.

As you can see from this mental exercise, for the $(n,n)$-categories there isn't really a space floating around, per se. Of course secretly there is a space, but it is hidden. This notion is much more like the classical notion of category and bicategory. Hope this helps!