This is trivial if $\gcd(\ell, p - 1) = 1$, and I assume it's easy otherwise, but I won't try to remove that restriction.
Recall Fermat's little theorem: $z^{p - 1} \equiv 1 \pmod p$. Let $\ell_{inv} = \ell^{-1} \pmod{p - 1}$, which exists as long as $\gcd(\ell, p - 1) = 1$.
Then, $(z^\ell)^{\ell_{inv}} \equiv z^{\ell \cdot \ell_{inv}} \equiv z^{k (p - 1) + 1} \pmod p$, for some $k \in \mathbb{Z}$, since $\ell \cdot \ell_{inv} \equiv 1 \pmod{p - 1}$.
Going on, $z^{k (p - 1) + 1} \equiv z^{k (p - 1)} z \equiv (z^{p - 1})^k z \pmod p$.
But by FLT, $z^{p - 1} \equiv 1 \pmod p$, so $(z^{p - 1})^k z \equiv 1^k z \equiv z \pmod p$. Thus, you have recovered $z$.
Numerical example in PARI/GP:
\\ Generation
? p = nextprime(random(2^31))
%1 = 1546275799
? z = random(p)
%2 = 879788114
? l = random(); while(gcd(l, p-1) != 1, l = random()); l
%3 = 1247963869
? zl = lift(Mod(z,p)^l)
%4 = 815723813
\\ Attack given only l, zl, p
? linv = lift(Mod(l, p-1)^(-1))
%5 = 914277349
? z_rec = lift(Mod(zl, p)^linv)
%6 = 879788114
? z == z_rec
%7 = 1
Just for completeness: the inversion modulo $p - 1$ has $O(p^2)$ complexity using a binary gcd, and the exponentiation to recover $z$ has complexity $O(p^3)$ if you use any non-naïve exponentiation algorithm, even the simplest algorithms such as left-right or right-left exponentiation, and even if you use the schoolbook multiplication algorithm. Indeed, it's the same cost as computing $z^l \pmod p$ itself, so it's as easy to break as it is to generate the parameters themselves.
