Is there any way to reduce this huge mass ?

Yes. =)

Think about places where you have duplicate, or nearly duplicate code, and write a function that does it for you.

Also think about ways you could store data in a new matrix while looping through your matrix.

rough pseudo code:

answers = [[]]
for x in matrix:
    answers[x] = count_neighbors_at_x(matrix,x)

One crazy way of doing this is using complex numbers. Let me know if you want an explanation the whole thing takes only a couple of lines.

But a very simple way is to use a dictionary instead of a nested list, and then use sum() with a comprehension.

from random import randint 
values = { (r,c):randint(0, 9) for r in range(5) for c in range(5) }

def neighbors(values,x,y):
    return sum( values[x+dx, y+dy]
            for dx,dy in [(-1,-1),(-1,0),(-1,1),(0,-1),(0,1),(1,-1),(1,0),(1,1)] if (x+dx, y+dy) in values )

tot = sum( neighbors(values, x,y) for x,y in values )

print(values)
print(tot, tot/25)

If you don't know what a comprehension is, now's a fine time to learn. It's a shortcut for having loops and temporary values, and just returns the object instead.

 tot = sum(neighbors(values, x,y) for x,y in values)

is the same as:

tot = 0
for x,y in values:
    tot += neighbors(values, x,y)

It's just shorter and easier to follow once you understand it.

similarly

    return sum( values[x+dx, y+dy]
            for dx,dy in [(-1,-1),(-1,0),(-1,1),(0,-1),(0,1),(1,-1),(1,0),(1,1)] if (x+dx, y+dy) in values )

is the same as:

tot = 0
for dx,dy in [(-1,-1),(-1,0),(-1,1),(0,-1),(0,1),(1,-1),(1,0),(1,1)]:
    if  (x+dx, y+dy) in values):
        tot += v[x+dx, y+dy]