How strong is the gravity from half of the distant universe? I would like to know, how strong would the gravity be from the distant universe if we were somehow able to remove half of it.  For example, if everything behind us did not exist, how strongly would the stuff in front of us be pulling?  By distant, let us consider everything beyond 10 billion light years (e.g. remove all the closer stuff).  How many G's acceleration would I experience (discounting the fact that I would be in free fall and wouldn't feel it)?
Not very strong at all. I get a rough figure of $3.725\times 10^{-9} m/s^2$. 

To perform that calculation I made a few simplifying assumptions.


Assume that we can ignore everything outside the observable
universe.   
Assume that the observable universe is a static homogeneous sphere
with a radius of 46.6 billion light-years, and a mean density of
$\rho = 0.85\times 10^{-26} kg/m^3$, that figure is the critical
density
needed for the global curvature of the universe to be zero (i.e., on
the very large scale spacetime is flat).   
Assume that we can use Newtonian physics to calculate the
acceleration due to gravity.


I didn't bother removing all the stuff closer than 10 billion light years.

We're standing at the centre of the flat surface of a hemisphere of radius $r$ and we want to know how strong the gravity is. We can pretend that all the mass of the hemisphere is concentrated at its centre of mass, which is $\frac{3r}{8}$ below us.

The Newtonian formula for gravitational acceleration is 

$$a = \frac{GM}{r^2}$$

Mass is volume times density, i.e., $M=V \rho$ and the volume of a hemisphere is

$$V = \frac{2}{3}\pi r^3$$

Combining everything, not forgetting to use $\frac{3r}{8}$ as the distance term in the denominator of the acceleration formula, we get

$$a = G \rho \frac{128}{27} \pi r$$

Plugging in the values I mentioned earlier for $\rho$ and $r$, the Google Calculator says $3.72484086\times 10^{-9} m/s^2$. That's about $3.8\times 10^{-10}$ standard gravities.