You can use the inclusion–exclusion principle to count the number of ways to do the $k$ rounds so that each ball is picked at least once:
Denote by $X$ the set of all $k$-rounds without restrictions and denote by $A_i$ the set of all the $k$-rounds in which $i$-th ball isn't picked. We are looking for $e_0=|X\setminus(A_1\cup...\cup A_N)|$.
Denote $s_0=|X|=\binom{\binom{N}{n}+k-1}{k}, \hspace{10pt} s_j=\sum_{1\leq i_1<... We have $|A_i|=\binom{\binom{N-1}{n}+k-1}{k}, \hspace{10pt} s_1=\binom{N}{1}\binom{\binom{N-1}{n}+k-1}{k}$ Since each time we picked $n$ balls out of $N-1$ (since we excluded the $i$-th), and we did this $k$ times, i.e. we count $k$-multisubsets of a set with $\binom{N-1}{n}$ elements.. To get $s_1$, we need to choose the $i$ and then multiply by the size of $A_i$, since the size of $A_i$ does not depend on $i$.
Similarly, $|A_i\cap A_j|=\binom{\binom{N-2}{n}+k-1}{k}, \hspace{10pt} s_2=\binom{N}{2}\binom{\binom{N-2}{n}+k-1}{k}$ Since we have to exclude two balls now. Continue in the same fashion to get: $s_j=\binom{N}{j}\binom{\binom{N-j}{n}+k-1}{k}$ Be the inclusion–exclusion principle, we have: $e_0=\sum_{j=0}^N (-1)^js_j=\sum_{j=0}^N (-1)^j\binom{N}{j}\binom{\binom{N-j}{n}+k-1}{k}$ Then the probability you were looking for is $\frac{e_0}{|X|}$.
Remark: Observe that the sum runs up to $N$. It is possible that you won't have enough balls at some point, but the binomial coefficient will vanish once $n>N-j$, so there is no need to fix the summation limits.