First, we find the probability that the cubes are positioned correctly to be able to provide as many black faces as necessary. There are 8 corner cubes, which must go at the 8 corners. There are 12 edge cubes, which must go at the 12 edges. There are 6 face-center cubes, which must go at the 6 face-centers. There is one center cube, which must go at the center. The probability is $\frac{8!\cdot 12!\cdot 6!\cdot 1!}{27!}$. Next, we find the probability that, if the cubes are positioned correctly, they are oriented correctly. The black corner of the corner cubes must face outward, so the probability is $(1/8)^8$. The black edge of the edge cubes must face outward, so the probability is $(1/12)^{12}$. The black face of the face-center cubes must face outward, so the probability is $(1/6)^6$. The center cube can be oriented any way. The complete probability is $\frac{8!\cdot 12!\cdot 6!\cdot 1!}{27!}\cdot\frac{1}{8^8 \cdot 12^{12} \cdot 6^6}$, which works out to be 1/5465062811999459151238583897240371200, roughly $1.83 \times 10^{-37}$.