Constructing a DFA
There is an algorithm to construct a minimal DFA for a regular language. In this case, it isn't too hard to come up with a DFA with $z(z+1)/2$ states.
Observe that $L_z = \{\epsilon, abc, aabbcc, \ldots, a^{z-1}b^{z-1}c^{z-1}\}$. Once you know the number of $a$'s at the beginning of the word, you know how many $b$'s and $c$'s to expect. Here's a simple DFA built from this observation (with $z' = z-1$):

[source]
Following the usual convention, if there is no transition for a particular letter from a particular node, the automaton goes to a sink state.
This automaton has $n$ states reached after reading $0$ to $z-1$ letters a
($k,0,0$ for $0\le k\le z-1$), and branches of length $2k$ for $0 \le k \le z-1$ to read the right number of b
's and c
's ($k,i,0$ and $k,k,i$ for $1 \le i \le k \le z-1$). That's a total of $z + \sum_{k=0}^{z-1} (2k) = z^2$ states, plus one to account for the implicit sink state.
It's possible to compress this automaton a bit.
Notice that every side branch finishes by counting the c
's. Instead of tracking how many c
's have already been seen, think in terms of how many c
's are left to read. You can merge all the states of the form $k,k,k-n$, for all $k \ge 1$, into a state $-,n$ from which $n$ occurrences of c
must be read. In particular, there will be only two final states: one for the empty word, and one after reading the right number of c
's. In this automaton, the $k$th side branch has $k$ edges to read the b
's (so $k-1$ states, as the start state belongs to the a
trunk, and the end state belongs to the c
collector), and there are $z-1$ transitions to count the c
's (so $z$ states, corresponding to $0$ through $z-1$ c
's left).
The total number of states in this automaton (including the sink state) is $z + \sum_{k=1}^{z-1} (k-1) + (z-1) + 1 = z(z+1)/2 + 1$.
Constructing an NFA
In other words, $n_z$ is the number of states in a minimal NFA that recognizes $L_z$. If you have a DFA with $N$ states, that's an NFA with $N$ states. So $n_z \le z(z+1)/2+1$. For most languages, a minimal NFA is smaller than a minimal DFA: nondeterminism adds expressive power in terms of how much you can accomplish with a number of states.
In this particular case, it happens that the minimal DFA above is also a minimal NFA (if I didn't make a mistake). Here's an informal proof.
Consider the states in the middle of reading the b
's. When reading $a^k b^i | b^{k-i} c^k$, where $|$ represents the current position, the state must account for both $k$ and $i$, to remember both the number of seen a
's (which is necessary, because we'll need to count the c
's later) and the number of seen b
's (or, equivalently, the number of b
's still required). This requires distinct states for every $(i,k)$ such that $1 \le i \lt k \le z-1$. In addition, there must be $z-1$ states to count the a
's at the beginning, and $z-1$ states to count the c
's at the beginning, plus a sink state. That's a total of $z(z+1)/2+1$ states.