Do I have a chance to get a closed form for this integral? I conjecture that
$$\int_{-z}^z \frac{\sin \left(\frac \pi {2z} x + \frac \pi 2 \right)} z \int_z^\infty \exp\left({- \frac {(y-x)^2 \pi^2}{16}}\right) \,d y d x  \sim \frac 1 {z^2}$$
when $z\to \infty$.
Maybe there should be also equality.
 A: Changing $x=zt, y=zt+u$, on can express
\begin{align}
 I&=\int_{-z}^z \frac{\sin \left(\frac \pi {2z} x + \frac \pi 2 \right)} z \int_z^\infty \exp\left({- \frac {(y-x)^2 \pi^2}{16}}\right) \,d y d x\\
 &=\int_{-1}^1\cos\frac{\pi t}{2}\,dt\int_{z(1-t)}^\infty \exp\left( -\frac{\pi^2}{16} u^2 \right)\,du
\end{align}
It can be integrated by parts,
\begin{align}
 I=\frac{2}{\pi}&\left[\sin \frac{\pi}{2}\int_{0}^\infty \exp\left( -\frac{\pi^2}{16}t^2 \right)\,dt \right. \\
 -&\sin \left( \frac{-\pi}{2} \right)\int_{2z}^\infty \exp\left( -\frac{z^2\pi^2}{16}t^2 \right)\,dt\\
  -&z\left.\int_{-1}^1\sin \frac{\pi t}{2}\exp\left( -\frac{z^2\pi^2}{16}(1-t)^2 \right)\,dt
  \right]
\end{align} 
or,
\begin{equation}
 I=\frac{2}{\pi}\left[\frac{2}{\sqrt{\pi}}\left( 2-\operatorname{erf}\left( \frac{z\pi}{2} \right) \right)-z\int_{0}^2\cos \frac{\pi s}{2}\exp\left( -\frac{z^2\pi^2}{16}s^2 \right)\,ds\right]
\end{equation} 
For $z\to\infty$, one can evaluate the integral using the Laplace method,
\begin{equation}
z\int_{0}^2\cos \frac{\pi s}{2}\exp\left( -\frac{z^2\pi^2}{16}s^2 \right)\,ds\sim \frac{2}{\sqrt{\pi}}-\frac{2}{\sqrt{\pi}z^2}+\frac{1}{\sqrt{\pi}z^4}+O\left( z^{-6} \right)
\end{equation} 
while (DLMF)
\begin{equation}
\frac{2}{\sqrt{\pi}}\left( 2-\operatorname{erf}\left( \frac{z\pi}{2} \right) \right)\sim \frac{2}{\sqrt{\pi}}+O\left( z^{-1}e^{-z^2\pi^2/4} \right)
\end{equation} 
and thus
\begin{equation}
I\sim \frac{4}{\pi^{3/2}}\frac{1}{z^2}\left[1-\frac{1}{2z^2}+O\left( z^{-4} \right)\right]
\end{equation} 
Alternatively, by expressing the $\cos$ in complex and completing the exponent, the integral can be expressed as
\begin{equation}
z\int_{0}^2\cos \frac{\pi s}{2}\exp\left( -\frac{z^2\pi^2}{16}s^2 \right)\,ds=\frac{\exp(-1/z^2)}{\sqrt{\pi}}\left[
\operatorname{erf}\left( \frac{z^2\pi+2i}{2z} \right)+\operatorname{erf}\left( \frac{z^2\pi-2i}{2z} \right)
\right]
\end{equation} 
which gives a closed form expression for the integral.
A: As Paul Enta already answered, the integral can be exactly computed.
$$J=\int_z^\infty \exp\left({- \frac { \pi^2}{16}(y-x)^2}\right) \,d y=\frac{2 }{\sqrt{\pi }}\left(1+\text{erf}\left(\frac{\pi}{4}   (x-z)\right)\right)$$
$$I=\int \frac{\cos \left(\frac{\pi  x}{2 z}\right)}{z}J \,dx=\frac{2 e^{-\frac{1}{z^2}}}{\pi ^{3/2}}\left(\text{erf}\left(\frac{\pi  z (z-x)+4 i}{4 z}\right)-\text{erf}\left(\frac{\pi  z (x-z)+4 i}{4 z}\right) \right)+\frac{4 \sin
   \left(\frac{\pi  x}{2 z}\right)}{\pi ^{3/2}} \left(1+\text{erf}\left(\frac{\pi  (x-z)}{4} \right)\right)$$ Now, using the bounds, the result of the given definite integral is
$$\frac{2 \left(4-2 \text{erf}\left(\frac{\pi  z}{2}\right)-i e^{-\frac{1}{z^2}}
   \left(\text{erfi}\left(\frac{1}{z}-\frac{i \pi 
   z}{2}\right)-\text{erfi}\left(\frac{1}{z}+\frac{i \pi 
   z}{2}\right)\right)\right)}{\pi ^{3/2}}$$ and expanding for large values of $z$, this gives
$$\sim \frac{4}{\pi ^{3/2} z^2}\left(1-\frac{1}{2 z^2}+\frac{1}{6 z^4}+O\left(\frac{1}{z^6}\right) \right)+\frac{32 e^{-\frac{1}{4} \pi ^2 z^2}}{\pi ^5 z^5}\left(1-\frac{12}{\pi ^2 z^2}+\frac{4 \left(45-\pi ^2\right)}{\pi ^4
   z^4}+O\left(\frac{1}{z^6}\right) \right)$$
