= Solution
The error $e(f)=f''(0)-f(-1)+2f(0)-f(1)$ is linear and annihilates all degree-at-most-two <polynomials>. The <Peano kernel theorem> with <derivative> order three gives
$$
e(f)=\int_{-1}^1K(t)f'''(t)\,dt,\qquad
K(t)=e_x\left(\frac{(x-t)_+^2}{2}\right).
$$
For $t\ne0$, the second <derivative> of the truncated quadratic at $x=0$ is $\mathbf1_{\{t<0\}}$. Its values at $x=-1,0,1$ then give
$$
K(t)=\mathbf1_{\{t<0\}}+(-t)_+^2-\frac{(1-t)^2}{2}
=\begin{cases}\tfrac12(1+t)^2,&-1<t<0,\\-\tfrac12(1-t)^2,&0<t<1.\end{cases}
$$
The value at the single point $t=0$ is irrelevant to the integral. This is the <centered second-derivative Peano kernel>. Its absolute integral is
$$
\int_{-1}^1|K(t)|\,dt
=\frac12\int_{-1}^0(1+t)^2dt+\frac12\int_0^1(1-t)^2dt
=\frac13.
$$
Consequently
$$
\boxed{|e(f)|\le\frac13\|f'''\|_{C[-1,1]}.}
$$
Its changing sign explains why cubic <polynomials> also have zero error: their third <derivative> is constant and $\int_{-1}^1K=0$. No fourth <derivative> is needed for the claimed bound.
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