How many colors in 3-d space to paint boxes? Let's imagine that we have boxes shaped as rectangular cuboids and colored with many different colors (one color for one box). 
Boxes can touch themselves by faces. Their edges are parallel to axes $\mathbf{x,y,z}$ (this is additional constraint for construction made of boxes).
What number of colors is needed to fulfill condition for cuboids not to touch by any face with the cuboid of the same color?
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For an arbitrary 3-d shapes it seems there is no upper limit for a number of colors,  is it any limit for cuboids? if not why ?
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Special cases show limited number of colors. If we have, for example, boxes with equal height  (but different other dimensions) and start to built construction from the plane $Oxy$ then we can build layers of boxes of the same height and the first layer can be colored with 4 colors (as any planar map), the second one with extra 4 colors, a third one can be colored with 4 colors of the first layer, so the total number of colors for such expanding construction is only 8 at the worst case. 
And what happens when we allow that boxes have limited range of heights, let's say $h, 2h,3h ..nh  $ (n-natural).
Can needed number of colors be expressed with some formula $c(n)$ ? Let's consider at least case for two values of height: $h$ and $2h$.
 A: Here's an answer for a slight generalization of the special case you gave:

[T1] Let $S$ be a finite set of nondegenerate disjoint axis-aligned cuboids in $\mathbb{R}^3$. Let $H > h > 0$ such that no cuboid is taller than $H$ units or smaller than $h$ units in height (the $z$-axis). Then it is possible to color the cuboids in $S$ using at most $4(\lceil H/h \rceil+2)$ colors such that any two cuboids touching by faces have different colors.

Proof: Assume wlog. that $h = 1$ and that $H$ is an integer (for noninteger $H$ replace $H$ by $\lceil H \rceil$). In this case, $4(\lceil H/h \rceil+2)$ simplifies to $4(H+2)$.
For each cuboid $C \in S$ consider the smallest axis-aligned cuboid $C'$ containing $C$ whose top and bottom faces have integer $z$-coordinates, let $S'$ be the set of all these cuboids. Observe that if any two cuboids $C, D \in S$ touch by faces, then $C'$ and $D'$ touch or intersect (1). Observe also that no two intersecting cuboids $C', D' \in S'$ have coplanar top or bottom faces (2).
Let the elevation $e(C)$ of any cuboid denote the minimum $z$-coordinate. By construction, all $C' \in S'$ have integer elevations, and integer heights not greater than $H+1$. The idea is now to group the set of cuboids into $H+2$ four-colorable subsets using elevation.
Define $S'_i := \{C' \in S'\mid e(C') = i \}$ for all $i \in \mathbb{Z}$. By (2), no two cuboids in $S'_i$ intersect or are situated above each other. Coloring the cuboids in $S'_i$ therefore reduces to coloring their bottom faces, which requires four colors as per the four-color theorem as these bottom faces are all coplanar.
Now observe that if we have $|i-j| > H+1$, any $C' \in S'_i, D' \in S'_j$ do not touch or intersect each other. Hence we can use four colors to color the set $$T'_i := \bigcup_{k \in \mathbb{Z}}S'_{(H+2)k+i}$$
and since $S' = T'_0 \cup \ldots \cup T'_{H+1}$, we can use $4(H+2)$ colors to color the set $S'$ so that any two cuboids touching or intersecting have distinct colors.
By (1), this coloring induces a valid coloring of $S$. $~~~~\square$
EDIT:
There's a much, much better approximation that follows from this.

[T2] Let $S$ be a finite set of $n>0$ nondegenerate disjoint axis-aligned cuboids in $\mathbb{R}^3$. Then it is possible to color the cuboids in $S$ using at most $16(1+\lfloor\log_2(n)\rfloor)$ colors such that any two cuboids touching by faces have different colors.

Proof: Let $a_0 < \ldots < a_k$ be all possible altitudes of all top and bottom faces of the cuboids in $S$, where $1 < k < 2n$ since we have only $2n$ available top and bottom faces. Choose a strictly increasing continuous surjective map $f: \mathbb{R} \to \mathbb{R}$ that maps $a_i$ to $i$ for all $i = 0, \ldots, k$. Construct the homeomorphism $F : \mathbb{R}^3 \to \mathbb{R}^3, (x, y, z) \mapsto (x, y, f(z))$. This map maps cuboids to cuboids, and two cuboids $C, D$ touch if and only if $F(C), F(D)$ touch.
We now let $S' := F(S)$. Now $S'$ is a set of cuboids whose top and bottom faces have integer altitudes between $0$ and $2n-1$, therefore each cube has an integer height between $1$ and $2n-1$.
We partition the set $\{1, \ldots, 2n-1\}$ into $p := 1+\lfloor\log_2(n)\rfloor$ subsets $A_1, \ldots, A_p$ such that $\max(A_i)/ \min(A_i) \leq 2$ for all $i = 1, \ldots, p$. Let $S'_i := \{C' \in S' \mid h(C') \in A_i\}$. By [T1], we can color each of these $S'_i$ using at most $16$ colors each. Since $S' = S'_1 \cup \ldots \cup S'_p$, there exists a coloring of $S'$ (therefore also of $S$) using at most $16p$ colors.
Note: We can easily improve this bound to $12p$ if we modify [T1] a little.
PS: I have deduced another upper bound for the amount of colors required that depends only on the maximum aspect ratio of the faces of all cuboids. For a maximum aspect ratio of $p$, one needs at most $\lfloor2p^2+12p+13\rfloor$ colors.
