How can the response/output of a given filter to a given sampled data be obtained in this case?
For discrete time, we typically use the Z-Transform.

It is essentially the discrete version of the Laplace Transform. But there had to be a representation for discrete samples. The main difference, other than using the Z-Transform as a discrete method, is that the Z-Transform as a radius of convergence (ROC), meaning when you perform the summation operation, your series will converge in a particular \$z\$ value.

The Z-Transform has similar properties to that of the Laplace Transform. We notate a function of the Z-Transform as something like \$f[n]\rightarrow F(z)\$.

So 

\$Y(z)=X(z)H(z)\$

If you wish, you can make \$x(t)\$ into \$x[n]\$ and the only difference is that you're saying it's a discrete function with infinitely many samples as \$\Delta n\ \rightarrow \infty\$. If you look at the Wikipedia hyperlink from the first sentence of my answer, you will see the table for Z-Transform along with its properties.

So getting back to your low pass filter. There are ways to essentially convert from Laplace Domain to the Z-Domain. This is typically converted allowing \$s=\frac{1}{T} \ln({z})\$, where \$T\$ is the sampling period. Filtering in the Z-Domain can get really messy.

When filtering in the Laplace Domain, we represent \$s\$ as \$j\omega\$... Well, in the Z-Domain, we represent \$z\$ as \$e^{j\omega}\$.