Obtaining normalized matrix for the Haar Wavelet Transform
After a few quick calculations, it seems to me that the trouble comes from poor notations for the root in your reference. If you read, in the final normalized matrix, $\sqrt{8/64}$ and $\sqrt{2/4}$ instead of  $\sqrt{8}/64$ and $\sqrt{2}/4$ (along with the $\pm$ signs), then  the final result is correct.

The matrices $V_i$ are orthogonal. To normalize them, you only need to multiply them by a diagonal matrix $D_i$ made from the inverse of the norm of each column. 

For $V_1$, the norms are all equal to $\sqrt{1/2^2+1/2^2}=\sqrt{1/2}$. Hence, you can multiply $V_1$ by the diagonal matrix
$$D_1= \operatorname{Diag} \{\sqrt{2},\,\sqrt{2},\,\sqrt{2},\,\sqrt{2},\,\sqrt{2},\,\sqrt{2},\,\sqrt{2},\,\sqrt{2}\}\,.$$
and get an orthonormal matrix. Similarly, you get that:
$$D_2= \operatorname{Diag} \{\sqrt{2},\,\sqrt{2},\,\sqrt{2},\,\sqrt{2},\,1,\,1,\,1,\,1\}\,,$$
and
$$D_3= \operatorname{Diag} \{\sqrt{2},\,\sqrt{2},\,1,\,1,\,1,\,1,\,1,\,1\}\,.$$

Finally, because of commutativity, you get a matrix $D=D_1 D_2 D_3$:

$$D= \operatorname{Diag} \{2\sqrt{2},\,2\sqrt{2},\,2,\,2,\,\sqrt{2},\,\sqrt{2},\,\sqrt{2},\,\sqrt{2}\}\,.$$

If you now multiply the matrix $W$ (the one with the $1/8$)



with that matrix $D$, you get a final matrix $W$:



provided the notation interpretation for the square root. For instance, you can recover $1/8\times 2\sqrt{2} = \frac{\sqrt{8}}{\sqrt{8^2}} =\sqrt{8/64} $.