I’ll do this for separate numbers $m$ of distinct coupons collected and $n$ of distinct coupons regarded as common because I don’t see any simplification allowed by the special case $m=n=N'$.
By linearity of expectation, the expected number of distinct rare coupons collected is the sum over $i\gt n$ of the probabilities $c_i$ of coupon $i$ being collected.
This is a generalization of Last coupon collected in the coupon collectors problem, which amounts to the special case $m=N-1$. We can solve the more general problem with the same Poissonization approach.
So let coupons be drawn in a continuous Poisson process with rate $1$ that is the sum of $N$ independent Poisson processes with rates $p_i$.
The probability density for the process for coupon $i$ to yield its first coupon at time $t$ is $p_i\mathrm e^{-p_it}$. At that time, the probability that $m$ distinct coupons have not yet been collected is
$$
\sum_{S\subset P_i}(-1)^{|S|-\ell}\binom{|S|-1}{\ell-1}\prod_{j\in S}\mathrm e^{-p_jt}\;,
$$
where $\ell=N-m$, $P_i$ is the set of sets of coupons other than $i$ with at least $\ell$ coupons, and $(-1)^{|S|-\ell}\binom{|S|-1}{\ell-1}$ is minus the Möbius function on $P_i\cup\{\emptyset\}$, ordered by inclusion, which you can check from
$$
\sum_{T\supset S\subset P_i}(-1)^{|S|-\ell}\binom{|S|-1}{\ell-1}=\sum_{\ell\le k\le|T|}(-1)^{k-\ell}\binom{k-1}{\ell-1}\binom{|T|}{k}=1\;.
$$
Thus, the probability $c_i$ that when coupon $i$ is first collected $m$ distinct coupons have not yet been collected, and thus coupon $i$ gets collected, is
\begin{eqnarray}
c_i
&=&
\int_0^\infty\mathrm dtp_i\mathrm e^{-p_it}\sum_{S\subset P_i}(-1)^{|S|-\ell}\binom{|S|-1}{\ell-1}\prod_{j\in S}\mathrm e^{-p_jt}
\\
&=&
\sum_{S\subset P_i}(-1)^{|S|-\ell}\binom{|S|-1}{\ell-1}\frac{p_i}{p_i+\sum_{j\in S}p_j}\;.
\end{eqnarray}
For $m=N-1$ and thus $\ell=1$, we recover the probability for coupon $i$ not to be the last coupon collected,
$$
c_i=\sum_{S\subset P_i}(-1)^{|S|-1}\frac{p_i}{p_i+\sum_{j\in S}p_j}\;.
$$
For $m=1$ and thus $\ell=N-1$, the sum has a single term, in which $S$ is the set of all $\ell$ coupons other than $i$, and we recover the trivial result $c_i=p_i$.
The expected number of rare coupons collected is
$$
\sum_{i\gt n}c_i=\sum_{i\gt n}\sum_{S\subset P_i}(-1)^{|S|-\ell}\binom{|S|-1}{\ell-1}\frac{p_i}{p_i+\sum_{j\in S}p_j}\;.
$$
We can write this as a single sum by collecting the terms with the same denominator:
$$
\sum_{i\gt n}c_i=\sum_T(-1)^{|T|-\ell-1}\binom{|T|-2}{\ell-1}\frac{\sum_{n\lt j\in T}p_j}{\sum_{j\in T}p_j}\;,
$$
where the sum now runs over all sets of at least $\ell+1$ coupons.
I don’t see how either of these representations could be simplified for the specific case of a Zipf distribution.
