How to find the derivative value at $(\pi,0)$ for this implicit function $n$ times?

You can substitute a Taylor expansion in your equation

y=Sum[a[n](x-π)^n, {n,1,6}]+ O[x,π]^7

SolveAlways[ Sin[x+y]+Sin[x]-y ==0, x]

(* {{a[5]->-(1/240),a[6]->0,a[4]->0,a[3]->1/12,a[2]->0,a[1]->-1}} *)

Last[Nest[Function[d, {d, Append[Last[#], Solve[0 == d /. x -> π /. Last[#]][[1, 1]]]}][
                   D[First[#], x]] &, {Sin[x] + Sin[x + y[x]] - y[x], {y[π] -> 0}}, 15]]

$$\left\{y(\pi )\to 0,y'(\pi )\to -1,y''(\pi )\to 0,y^{(3)}(\pi )\to \frac{1}{2},y^{(4)}(\pi )\to 0,y^{(5)}(\pi )\to -\frac{1}{2},y^{(6)}(\pi )\to 0,y^{(7)}(\pi )\to \frac{1}{2},y^{(8)}(\pi )\to 0,y^{(9)}(\pi )\to 17,y^{(10)}(\pi )\to 0,y^{(11)}(\pi )\to -\frac{1153}{4},y^{(12)}(\pi )\to 0,y^{(13)}(\pi )\to \frac{13297}{4},y^{(14)}(\pi )\to 0,y^{(15)}(\pi )\to \frac{108109}{2}\right\}$$

You need to find the Taylorpolynom for your function?

The code for the Taylor formula:

taylor = (vars - point).# &;
init := D[f[vars], {vars, j}] /. Thread[vars -> point];
taylorPolynom[m_] := Sum[1/j! Nest[taylor, init, j], {j, 0, m}]

vars = {x, y};
point = {Pi, 0};
f[vars_] = Sin[x + y] + Sin[x] - y;

taylorPolynom[2]
-2 (-Pi + x) - 2 y

or

taylorPolynom[6] // FullSimplify
1/120 (2 (120 + (-20 + (Pi - x)^2) (Pi - x)^2) (Pi - x) - 
   5 (48 + (-12 + (Pi - x)^2) (Pi - x)^2) y + 
   10 (-6 + (Pi - x)^2) (Pi - x) y^2 - 
   10 (-2 + (Pi - x)^2) y^3 + 5 (Pi - x) y^4 - y^5)