Update 18.10.2017: In the previous version of my answer I'd assumed the series begins at $x_0$ ; instead of $-x_1$ as it was defined in the question. I adapted now my results and matrices accordingly.
Giving examples to questions in the comments.
I used Pari/GP; there is a Cesaro-sum-compatible procedure in it sumalt()
With this I got the following table:
f(x)=sqrt(x+1/2)
p=4 \\ used as exponent for the series
list=vectorv(20) \\ takes 20 solutions
\\ --------- put the following commands of the loop in a "bracketed block"
{ for(q=1,20, x0=x1=q-1;
su=sumalt(k=1,
(-1)^k * ( x1 = f(x1))^p); \\ for iteration (k>0)
list[q] = [ x0 , su , 1.0/8 - x0^2 ]
); }
\\ ---------
printp(Mat(list));
...
x0 ! sum by"sumalt" ! 1/8 -x0^2
! (cesarosum) ! equals "sum"
-----+ ----------------+ ------------------
0 0.125000000000 0.125000000000
1 -0.875000000000 -0.875000000000
2 -3.87500000000 -3.87500000000
3 -8.87500000000 -8.87500000000
4 -15.8750000000 -15.8750000000
5 -24.8750000000 -24.8750000000
6 -35.8750000000 -35.8750000000
7 -48.8750000000 -48.8750000000
8 -63.8750000000 -63.8750000000
The differences are in the digits near software-epsilon
[Update] There are some more examples of interesting sums, see the following answer in
another MSE-thread
There is also an empirical/a heuristic/conjectured result using my earlier discussed matrix-approach, which employs Carleman-matrices and for the summation of the alternating iteration series -when this is possible- the Neumann-series of that Carleman matrix.
My Pari/GP-tools give me the following
pc_f=polcoeffs(f(x),32)~ \\ put the (leading) coefficients of the
\\ powerseries-expansion of f(x) into a vector.
\\ Because of finite size we can always -at best-
\\ get approximative solutions
\\ show the first few coefficients:(actually I work with first 32 coeffs.)
[0.707106781187, 0.707106781187, -0.353553390593, 0.353553390593,
-0.441941738242, ... ]
F = mkCarlemanmatrix(pc_f) \\ user defined procedure
\\The top-left of F is
1 0.707106781187 0.500000000000 0.353553390593 0.250000000000 0.176776695297
0 0.707106781187 1.00000000000 1.06066017178 1.00000000000 0.883883476483
0 -0.353553390593 0 0.530330085890 1.00000000000 1.32582521472
0 0.353553390593 0 -0.176776695297 0 0.441941738242
0 -0.441941738242 0 0.132582521472 0 -0.110485434560
0 0.618718433538 0 -0.132582521472 0 0.0662912607362
With a vector $\small V(x)=[1,x,x^2,x^3,...]$ we can then evaluate the series by doing the dotproduct
V(x) * F = [1, f(x), f(x)^2, f(x)^3, f(x)^4 , ... ]
= V(f(x))
Here the series which occur at odd indices ($\small f(x),f(x)^3,f(x)^5$) have convergence-radius $\small \rho<=1$
Note, that in the fifth column (=at column with index 4) we get the 4'th power of the function: $\small f(x)^4$ which is of course what shall interest us below for our problem.
Now we try to get a meaningful matrix A by the Neumann-Series, which interprets the alternating geometric-series for matrices (as far as this is at all possible). Because your series begins at $x_1$ and not $x_0$ I omit the first term which would be the identity-matrix $F^0$:
$$\small A = - F+ F^2 - F^3 ... = F*(I + F)^{-1} $$
That matrix-inversion must be done with much care; in such cases as here I use LDU-decomposition, (exact(!)) inversion of the components and construction of the inverse from the product of inverses of the L,D,U components applying Eulersummation when convergences in the dotproducts are bad.
But even the naive inversion-procedure in Pari/GP gives a seemingly meaningful approximate solution:
A = -F * (matid(32) + F)^-1
\\ the top-left of A
-1/2 -1.3477094 1.0977094 -0.57076948 0.12500000 0.11653321
0 -1.0048085 0.0048084741 -0.0024627863 0 0.0013720680
0 9.3831760 -9.3831760 4.4965232 -1.0000000 -0.94229215
0 -0.36697128 0.36697128 -1.1877337 0 0.10440858
0 -5.4154218 5.4154218 -3.2908264 0 0.49966112
0 -3.9408033 3.9408033 -2.0077436 0 0.11044718
0 2.4551648 -2.4551648 1.4308645 0 -1.1883271
0 1.5382960 -1.5382960 0.88397114 0 -0.55144081
0 -3.4084780 3.4084780 -1.6751387 0 0.95787115
0 -2.8478962 2.8478962 -1.3608389 0 0.80723231
The dotproducts with a $V(x)$-vector should give approximately:
$$ V(x) \cdot_{\mathfrak E} A = [ a_0 , a_1(x), a_2(x) , a_3(x), a_4(x), ... ] $$
The ${\mathfrak E}$ means here that possibly Eulersummation is involved as far as the occuring summations are not convergent, or converge only badly.
But we have two interesting columns here: they have only finitely many
entries so that the alternating series for that exponents might be computable by finite polynomials:
column represents gives value
by evaluation
-----------------------------------------------------------------
a_0 = - x_1^0 + x_2^0 - ... + ... = -1/2
a_4(x) = - x_1^4 + x_2^4 - ... + ... = 1/8 -1*x^2
The value for $\small a_0$ complies with the evaluation of $\small -1+1-1 \ldots$ by Eulersummation and the values for $\small a_4(x)$ comply with the results gotten by series-summation $\small -x_1^4 + x_2^4 - ... + ...$
Such heuristics by the Neumann-matrices can be found in many places in tetration and iteration-series, however I had never time and energy to sit down and do the formal proofs for that concluded properties....