**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 connected graph has at least $c-1$ 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, the image has size $n^{d-1}$, and the size of the image is also at most $c - (c-1)/d$ since at least one vertex on each of those edges is redundant. So,

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

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

For $d=3$, $c \ge \frac 32 n^2 - \frac12$ (mentioned by Gjergji Zaimi in the comments). This is 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 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 $\frac 32 n^2 + \frac 32 n - 4$.

Therefore,

$$ \frac 32 n^2 - \frac 12 \le \min |C_3(n)| \le \frac 32 n^2 + \frac 32 n - 4.$$