(14) For the cube $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, the centers of faces $A B C D$ and $A D D^{\prime} A^{\prime}$ are $M$ and $N$, respectively. The sine value of the angle formed by the skew lines $M N$ and $B D^{\prime}$ is $\qquad$
$\qquad$.
(14) Let the edge length of a cube be 1, and establish a spatial rectangular coordinate system with $\overrightarrow{D A} 、 \overrightarrow{D C} 、 \overrightarrow{D D^{\prime}}$ as the basis, then $B(1,1,0), D^{\prime}(0, .0,1), M\left(\frac{1}{2}, \frac{1}{2}, 0\right)$, $N\left(\frac{1}{2}, 0, \frac{1}{2}\right)$, therefore $\overrightarrow{B D^{\prime}}=(-1,-1,1), \overrightarrow{M N}=\left(0,-\frac{1}{2}, \frac{1}{2}\right)$.
Since $\overrightarrow{B D^{\prime}} \cdot \overrightarrow{M N}=1,\left|\overrightarrow{B D^{\prime}}\right|=\sqrt{3},|\overrightarrow{M N}|=\frac{\sqrt{2}}{2}$, we have,
$$
\begin{array}{l}
\cos \left\langle\overrightarrow{B D^{\prime}}, \overrightarrow{M N}\right\rangle=\frac{\overrightarrow{B D^{\prime}} \cdot \overrightarrow{M N}}{\left|\overrightarrow{B D^{\prime}}\right| \cdot|\overrightarrow{M N}|}=\frac{\sqrt{6}}{3}, \\
\sin \left\langle\overrightarrow{B D^{\prime}}, \overrightarrow{M N}\right\rangle=\frac{\sqrt{3}}{3} .
\end{array}
$$

Therefore, the answer is $\frac{\sqrt{3}}{3}$.