In Excel, you would just put, into cell B2 "=A1-B1" where B1 is a constant. So if I have a list of numbers in Column A, I can just autofill the formula in B2 down the column and get a new list.

May be I am not understanding the question fully, since I do not use Excel. But why not write

b1 = 99;
a1 = {1, 2, 3, 4};
b2 = a1 - b1
  (* {-98, -97, -96, -95} *)

It is perhaps also worth mentioning that the Excel case where A an B are two columns, and you wish a 3rd column C containing the differences on each row, the Mathematica equivalent is to just subtract the two lists, which is done component-wise:

In[1]:= {a, b, c} - {d, e, f}

Out[1]= {a - d, b - e, c - f}

Does this better clarify?

I think it will be much more clear if you simply give a small actual example of an input and output.

For example, if A1= {1,2,3,4} and B=99, what do you want B2 to be? Which is the result of your Excel "B=A1-B1" operation? I wrote A1 as a row here, simply to save vertical space. But pretend it is a column vector instead. I think a simple example would be much better than long explanation using just words in this case.

Yes, now it is much more clear. Since C={20-n1,c1-n2,c2-n3,c3-n4,c4-n5,c5-n6} is the same as

        C={20-n1, 20-n1- n2, 20-n1-n2- n3, 20-n1-n2-n3- n4, 20-n1-n2-n3-n4-n5,  20-n1-n2-n3-n4-n5-n6}

Then one way to do it is

A0 = {n1, n2, n3, n4, n5, n6};
B0 = 20;
C0 = B0 - Accumulate[A0]  (* the excel equation *)
(* {20 - n1, 20 -n1-n2, 20 -n1-n2-n3, 20-n1-n2-n3- n4, 20-n1-n2-n3-n4-n5, 20-n1-n2-n3-n4-n5-n6} *)

Using numbers

A0 = {1, 2, 3, 4};
B0 = 20;
C0 = B0 - Accumulate[A0] (*the excel equation*)
(*  {19, 17, 14, 10} *)