= Solution
The ideal $(f)$ is prime, since an irreducible element of the polynomial ring is prime. By the <Hilbert Nullstellensatz>, it is the full vanishing ideal of $Z(f)$. Its <Zariski tangent space> at a point $a$ is therefore the kernel of the single linear form $df_a$, and has dimension $n-1$ or $n$.
At least one partial derivative $f_j$ is nonzero. In characteristic zero this is immediate for a nonconstant polynomial. In characteristic $p$, if all partial derivatives vanished, all exponents would be multiples of $p$. Since an algebraically closed field is perfect, $f$ would be a $p$th power of a nonconstant polynomial, contradicting irreducibility.
The nonzero $f_j$ cannot vanish everywhere on $Z(f)$: the Nullstellensatz would give $f_j^r\in(f)$, hence $f\mid f_j$, contradicting $\deg f_j<\deg f$. Thus some point has $df_a\ne0$, and the minimum tangent-space dimension is
$$
\boxed{\dim Z(f)=n-1.}
$$
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