Gravitational field

In Newtonian gravity the gravitational potential is an additive scalar field, i.e. given two masses $M_1, M_2$ the total potential is given by: $$V(vec{r}) = -frac{GM_1}{||vec{r}-vec{r}_1||} - frac{GM_2}{||vec{r}-vec{r}_2||}$$ where $vec{r}_1, vec{r}_2$ are the position vectors of those masses and $vec{r}$ is the location at which you measure the potential.

In the case of a small mass, the contribution to the potential can be ignored (this is all that happens for small masses, in general nothing special happens for small masses).

Note: One shouldn't take $vec{r}=vec{r}_1$ or $vec{r}=vec{r}_2$ since then we would obtain a divergent (self-)energy. In that case only one of the two terms should be taken into account.

The principle of superposition works for any number of masses: the total gravitational field is the vector sum of the gravitational fields from all the individual masses. Just adding the forces as vectors works fine for gravity almost everywhere away from black holes. You will find this true for electric and magnetic fields too.

If a huge mass is taken then the gravitational field gets affected, but how does the field actually gets altered, does it weaken or what?

It depends on the vector sum. If it is a huge chunk of mass, of course, the field strength is increased. If you carve out a hole at the core of some planet, you are equivalently reducing the mass and its strength. As noted, the field is just vector sum.

Newton used this to prove that the gravitational field outside a solid sphere was the same as if all the mass were at the center by imagining the solid sphere to be composed of many small masses.

We know that for a gravitational field calculation we consider a test mass, i.e. a small mass

It doesn't have to be small at all. Applying Newton second law for gravity we get :

$$ ma = G frac {Mm}{r^2} $$

Body mass $m$ cancels each other in both equation sides, thus we get centripetal acceleration due to gravity :

$$ a = G frac {M}{r^2} $$

Thus it not depends on body mass $m$ at all.