= Solution
For each $x\in\mathbb F_p$, count the roots of $y^2=x^3+x^2+x+1$, and add the identity at infinity. A nonzero <quadratic residue> gives two roots, zero gives one, and a nonsquare gives none. Equivalently, with the <Legendre symbol> extended by $\chi_p(0)=0$,
$$
\#\widetilde E_p(\mathbb F_p)=p+1+\sum_{x\in\mathbb F_p}\chi_p(x^3+x^2+x+1).
$$
The numbers of affine $y$ values, in increasing $x$ order starting at zero, are
$$
\begin{array}{c|l|c}
p&\text{root counts for }x=0,\ldots,p-1&\#\widetilde E_p(\mathbb F_p)\\\hline
3&(2,2,1)&6\\
5&(2,2,1,1,1)&8\\
7&(2,2,2,0,2,2,1)&12\\
11&(2,2,2,0,0,0,0,2,0,0,1)&10
\end{array}
$$
so the answer is
$$
\boxed{(\#\widetilde E_3,\#\widetilde E_5,\#\widetilde E_7,\#\widetilde E_{11})=(6,8,12,10).}
$$
All four reductions are <smooth algebraic curves>: the original PDF gives the <elliptic-curve discriminant> $-2^8$, so these odd primes are primes of <good reduction>.
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