First poisons ratio for isotropic materials applies equally between each pair of dimensions. That means that if the axial length is stretched $1\%$ both of the lateral dimensions would reduce by $\nu 1\%$. I don't know where you're getting the 0.25 number from.
Second, this approximation only works when the strain is small. Lets say we have an orthogonal parallelepiped with dimensions $x$, $y$, and $z$. If we apply a stress $\sigma_x$ streching the x direction and let the y and z dimensions be stress free. Then we'll have:
$$\epsilon_x=\frac{\sigma_x}{E}$$
$$\epsilon_y=-\nu\frac{\sigma_x}{E}$$
$$\epsilon_z=-\nu\frac{\sigma_x}{E}$$
So the new dimensions would be:
$$x'=x(1+\epsilon_x)$$
$$y'=y(1-\nu\epsilon_x)$$
$$z'=z(1-\nu\epsilon_x)$$
And the new volume would be:
$$V'=x'\,y'\,z'= x\,y\,z\, (1+\epsilon_x)(1-\nu\epsilon_x)^2$$
So the ratio of the new volume to the old volume would be:
$$\frac{V'}{V}=(1+\epsilon_x)(1-\nu\epsilon_x)^2=1+(1-2\nu)\epsilon_x+(\nu^2-2\nu)\epsilon_x^2+\nu^2\epsilon_x^3$$
To get the increase in volume, you can subtract the 1. Then there are three more terms, but since $\epsilon_x$ is assumed to be small then the square and cube terms should be dominated by the linear term. Thus those terms are usually neglected and you're left with $(1-2\nu)\epsilon_x$. If you have an incompressible solid then this value should always be zero, so then $\nu$ must equal $0.5$
Let us look at a more general example:
$$V=V'$$
Is really saying:
$$\begin{split}1=(1+\epsilon_x)(1+\epsilon_y)(1+\epsilon_z)=&1+ \\
& \epsilon_x+\epsilon_y+\epsilon_z + \\
& \epsilon_x \epsilon_y + \epsilon_y \epsilon_z + \epsilon_x \epsilon_z + \\ & \epsilon_x \epsilon_y \epsilon_z
\end{split}$$
but again with a assumption that the strains are small, the product of multiple strains should be really small, so this equation simplifies to:
$$0=\epsilon_x+\epsilon_y+\epsilon_z$$
Now the general equations for stain in isotropic materials are:
$$\epsilon_x=\frac1{E}(\sigma_x-\nu(\sigma_y+\sigma_z))$$
$$\epsilon_y=\frac1{E}(\sigma_y-\nu(\sigma_x+\sigma_z))$$
$$\epsilon_z=\frac1{E}(\sigma_z-\nu(\sigma_x+\sigma_y))$$
We can sum these all together to get:
$$0=\epsilon_x+\epsilon_y+\epsilon_z=\frac1{E}(1-2\nu)(\sigma_x+\sigma_y+\sigma_z)$$
So we see that one again to conserve volume $\nu=\frac12$
Now why didn't this work in your example?
When you applied the stretching both other directions should have contracted. In order for the second direction to not contract, that direction would also have have to be pulled. This pulling on the second direction would further contract the third direction, but it would also contract the first direction, so in order for that to stay at it's original displacement you'd have to pull even harder.
Then you specified that $\nu=-\frac{\epsilon_z}{\epsilon_x}$ but this is only true if there is only stress in the x direction which is not the case for your scenario (because there must have been a stress applied to keep the 4cm side at 4cm).
Finally the strain you applied was too large. Even if you let the two lateral dimensions strain freely, you'd get a Poisson's ratio not equal to .5 just because the small strain approximation would no longer be valid:
$$\epsilon_x=\frac{11}{7}-1\approx 0.57$$
to conserve volume:
$$\epsilon_y=\epsilon_z=\sqrt{\frac{7}{11}}-1 \approx -0.20$$
Which is quite a bit less than half.
