show/hide this revision's text 3 Added upper bound

The other answers have given constructions hence upper bounds. Here is a lower

Lower bound.:

Let there be $c$ cubes in a blocking configuration. Consider the graph whose vertices are cubes so that cubes are connected if they share a face. Any cube must be connected in graph has at least one of the $d$ coordinate directions to another cube. Ignore any connections except those in a spanning tree (treating the cubes as vertices), which has $c-1$ edges. By the pigeonhole principle, there is at least one direction with at least $(c-1)/d$ edges in that direction. When you project parallel to that axis, there are the image has size $n^{d-1}$ images, n^{d-1}$, and this the size of the image is of size also at most $c - (c-1)/d$. c-1)/d$ since at least one vertex on each of those edges is redundant. So,

For $d=3$, $c \ge \frac 32 n^2 - \frac12$ (mentioned by Gjergji Zaimi in the comments), which . This is quite close sharp for $n=3$ by the construction with $13$ cubes shown by Joel David Hamkins.

Upper bound ($d=3$):

Here is a construction of a connected blocking configuration with $\frac32n^2 + O(n)$ cubes related to Zack Wolske's construction constructions. We'll identify the cubes with lattice points. We start with the points $\lbrace(x,y,z) | x-y \equiv z \mod n, 0 \le x,y,z \lt n \rbrace$, illustrated for $n=7$.

|......X|  |x......|  |.x.....|  |..x....|  |...x...|  |....x..|  |.....x.||.....X.|  |......X|  |x......|  |.x.....|  |..x....|  |...x...|  |....x..||....X..|  |.....X.|  |......X|  |x......|  |.x.....|  |..x....|  |...x...||...X...|  |....X..|  |.....X.|  |......X|  |x......|  |.x.....|  |..x....||..X....|  |...X...|  |....X..|  |.....X.|  |......X|  |x......|  |.x.....||.X.....|  |..X....|  |...X...|  |....X..|  |.....X.|  |......X|  |x......||X......|  |.X.....|  |..X....|  |...X...|  |....X..|  |.....X.|  |......X|
This is made of two triangles of points, in the planes $x-y=z$ (marked X) and $x-y=z-n$ (marked x), which are separated by some distance. We'll first make the connections within the triangles, and then connect the triangles to each other.
To connect the bottom triangle, use $n-1$ extra points (marked A) to connect the base of the triangle in the plane $z=0$ to itself and to the points in the triangle with $z=1$. Then for $i = 2, ..., n-1$, use $\lceil (n-i-1)/2 \rceil$ points (marked O) to connect the points in the plane $z=i$ to each other and to the lower points in the triangle. These points have $y$ odd, and are just above an X in the layer below. 
|......X|  |x......|  |.x.....|  |..x....|  |...x...|  |....x..|  |.....x.||.....XA|  |......X|  |x.....O|  |.x.....|  |..x....|  |...x...|  |....x..||....XA.|  |.....X.|  |......X|  |x......|  |.x.....|  |..x....|  |...x...||...XA..|  |....X..|  |....OX.|  |.....OX|  |x.....O|  |.x.....|  |..x....||..XA...|  |...X...|  |....X..|  |.....X.|  |......X|  |x......|  |.x.....||.XA....|  |..X....|  |..OX...|  |...OX..|  |....OX.|  |.....OX|  |x.....O||XA.....|  |.X.....|  |..X....|  |...X...|  |....X..|  |.....X.|  |......X|
This has added $n-1$ A's (this is one of the few correct uses of apostrophes to indicate a plural), and $0+1+1+2+2+...+\lfloor (n-1)/2 \rfloor = \lfloor (n-1)^2/4 \rfloor$ O's (see A002620). 
We repeat the process upside down to connect the upper left triangle using $n-2$ A's and $\lfloor (n-2)^2/4 \rfloor$ O's. 
|......X|  |x......|  |.x.....|  |..x....|  |...x...|  |....x..|  |....Ax.||.....XA|  |O.....X|  |xO....O|  |.xO....|  |..xO...|  |...x...|  |...Ax..||....XA.|  |.....X.|  |......X|  |x......|  |.x.....|  |..x....|  |..Ax...||...XA..|  |....X..|  |....OX.|  |O....OX|  |xO....O|  |.x.....|  |.Ax....||..XA...|  |...X...|  |....X..|  |.....X.|  |......X|  |x......|  |Ax.....||.XA....|  |..X....|  |..OX...|  |...OX..|  |....OX.|  |.....OX|  |x.....O||XA.....|  |.X.....|  |..X....|  |...X...|  |....X..|  |.....X.|  |......X|
Finally, we connect the upper and lower triangles with $n-2$ Z's. There are many choices for how to do this. We'll put them in $z=1$, $y=n-2$, $1 \le x \le n-2$. 
|......X|  |x......|  |.x.....|  |..x....|  |...x...|  |....x..|  |....Ax.||.....XA|  |OZZZZZX|  |xO....O|  |.xO....|  |..xO...|  |...x...|  |...Ax..||....XA.|  |.....X.|  |......X|  |x......|  |.x.....|  |..x....|  |..Ax...||...XA..|  |....X..|  |....OX.|  |O....OX|  |xO....O|  |.x.....|  |.Ax....||..XA...|  |...X...|  |....X..|  |.....X.|  |......X|  |x......|  |Ax.....||.XA....|  |..X....|  |..OX...|  |...OX..|  |....OX.|  |.....OX|  |x.....O||XA.....|  |.X.....|  |..X....|  |...X...|  |....X..|  |.....X.|  |......X|
In total, this configuration contains $n^2$ X's, $2n-3$ A's, $n-2$ Z's, and $\lfloor (n-1)^2/4\rfloor +\lfloor(n-2)^2/4 \rfloor = {n-1 \choose 2}$ O's, a total of $\frac 32 n^2 + \frac 32 n - 5$ for some n4$. 
Therefore, 
$$ \frac 32 n^2 - \frac 12 \le \min |C_3(n)| \le \frac 32 n^2 + \frac 32 n - 4.$$
    Post Undeleted by Douglas Zare
show/hide this revision's text 2 fixed constant; added 45 characters in body

The other answers have given constructions hence upper bounds. Here is a lower bound.

Let there be $c$ cubes in a blocking configuration. Any cube must be connected in at least one of the $d$ coordinate directions to another cube. Ignore any connections except those in a spanning tree (treating the cubes as vertices), which has $c-1$ edges. By the pigeonhole principle, there is at least one direction with at least $(c-1)/d$ cubes connected edges in that direction. When you project parallel to that axis, there are $n^2$ n^{d-1}$ images, and this image is of size at most $c - (c-1)/d$. So,

$$ n^2 n^{d-1} \le c - \frac{c-1}{d} \le = c \frac{d-1}{d}$$ frac{d-1}{d} + \frac 1d$$

$$ c \ge \bigg(\frac {d}{d-1}\bigg)n^2. d}{d-1}\bigg)n^{d-1} - \frac {1}{d-1}. $$

For $d=3$, $c \ge \frac 32 n^2$n^2 - \frac12$ (mentioned by Gjergji Zaimi in the comments), which is quite close to Zack Wolske's construction of $\frac 32 n^2 + \frac 32 n - 5$ for some n.

    Post Deleted by Douglas Zare
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