What is the Von Neumann entropy of $rho = sum_ip_i|iranglelangle i| otimes rho_i$?

Operator $rho$ is not a tensor product, it's a sum of tensor products $$ p_1|1ranglelangle 1| otimes rho_1 + p_2|2ranglelangle 2| otimes rho_2 + dots + p_d|dranglelangle d| otimes rho_d. $$ This is not the same as $$ big(sum_ip_i|iranglelangle i|big) otimes big(sum_irho_ibig), $$ so your expansion isn't correct.

Also in general $S(A+B)neq S(A)+S(B)$, but in this situation the supports of $|i ranglelangle i|otimes rho_i$ and $|j ranglelangle j|otimes rho_j$ are orthogonal, so we can write $$ S(rho) = S(p_1|1ranglelangle 1| otimes rho_1) + dots + S(p_d|dranglelangle d| otimes rho_d) $$ Here $p_i|1ranglelangle 1| otimes rho_i$ is not a density matrix because it's scaled, i.e. its trace equals $p_i<1$, so technically $S$ is not defined. But for such matrices we also can define expression $S(M) = -sum_i lambda_itext{ln}lambda_i$, where $lambda_i$ are eigenvalues of $M$. It's easy to check that for $c>0$ and density matrix $rho$ we have $S(crho) = cS(rho) - ctext{ln}c$. So we can write $$ S(p_i|iranglelangle i| otimes rho_i) = p_iS(|iranglelangle i| otimes rho_i)-p_itext{ln}p_i = $$ $$ = p_ibig(S(|iranglelangle i|) + S(rho_i)big)-p_itext{ln}p_i = p_iS(rho_i)-p_itext{ln}p_i $$ After the summation we will have $$ S(rho) = sum _i p_iS(rho_i) - sum_ip_itext{ln}p_i = sum _i p_iS(rho_i) + S(overline{p}) $$