You can write all the limits as inequalities and then deduce the new bounds from them in order from outside to inside.
In your case, you have
$$
-1\lt x\lt1\;,\\
0\lt y\lt\sqrt{1-x^2}\;,\\
0\lt z\lt \frac y2\;.
$$
For the new integration order, $z$ is on the outside, so you need the overall bounds for $z$. From the given inequalities, the range of $x$ is $[-1,1]$, thus the range of $y$ is $[0,1]$ and thus the range of $z$ is $[0,\frac12]$.
$x$ is next, so we need to solve the second inequality for $x$, yielding $|x|\lt\sqrt{1-y^2}$. Together with $y\gt 2z$ from the third inequality, this yields $|x|\lt\sqrt{1-4z^2}$, so the range of $x$ is $[-\sqrt{1-4z^2},\sqrt{1-4z^2}]$. Finally, $y$ is bounded both by $y\gt2z$ and by $y\lt\sqrt{1-x^2}$, so its range is $[2z,\sqrt{1-x^2}]$. Thus the integral is
$$
\int_0^{\frac12}\int_{-\sqrt{1-4z^2}}^{\sqrt{1-4z^2}}\int_{2z}^{\sqrt{1-x^2}}f(x,y,z)\,\mathrm dy\,\mathrm dx\,\mathrm dz\;.
$$
You need to consider the domain of integration. This is a triangle in the $z-s$ plane, bounded by $s=t$, $z=T$ and $z=s$. When you exchange integrals you need to re-describe this region so the outer integral limits do not depend on the inner integral variable of integration. Specifically: $$\int_t^T\Big(\int_s^T f(z, s) dz\Big) ds=\int_t^T\Big(\int_t^z f(z, s) ds\Big) dz.$$ I hope this is enough for you to complete the integral yourself.
Best Answer
The region of integration is a triangle. Crudely drawn here:
To reverse the order, $y$ would run from $0$ to $4$.
Within that, $x$ would at first run from $y/2$ to $y$. But halfway up, $x$ would start running from $y/2$ to $2$. Most people would break up the integral in two: $$\int_{y=0}^2\int_{x=y/2}^y+\int_{y=2}^4\int_{x=y/2}^2$$
You could express it as a single integral like $$\int_{y=0}^4\int_{x=y/2}^{f(y)}$$ where $f$ is a piecewise function that changes behavior at $y=2$. $$f(y)=\begin{cases}y/2&y\leq2\\2&y>2\end{cases}$$ You can be "clever" and find this way to express the same function $$f(y)=\frac{y+2-|y-2|}{2}$$ So you have $$\int_{y=0}^4\int_{x=y/2}^{\frac{y+2-|y-2|}{2}}$$