Yes, it is a consequence of Bézout's theorem and the Stein factorisation.

Let $d_i$ be the degree of $f_i$ for $i = 1, \ldots, n$. Consider the vector space 
$$
S = \bigoplus_{i = 1}^n \mathbb{C}[z_0, \cdots, z_n]_{d_i}
$$
as complex algebraic variety, where $\mathbb{C}[z_0, \cdots, z_n]_{d_i}$ is the space of homogeneous polynomials of degree $d_i$. 
There are tautological sections $g_i\in \Gamma(S, \mathcal{O}_S[z_0, ..., z_n]_{d_i})$ for $i = 1,\ldots, n$. Let $\mathcal{X} = \mathrm{V}_+(g_1, \ldots, g_n)\subseteq \mathbb{P}^n_S$
 be the universal family over $S$. The projective variety in question can be identified with its fibre $\mathcal{X}_f$ at $f = (f_1, \ldots, f_n)\in S(\mathbb{C})$. 
By the Stein factorisation, the projection $\mathcal{X}\to S$ can be factorised as $\mathcal{X}\xrightarrow{\alpha} \bar{\mathcal{X}}\xrightarrow{\beta} S$ such that $\alpha$ is proper with connected fibres and $\beta$ is finite. For a generic $\mathbb{C}$-point $s\in S(\mathbb{C})$, the fibre $\mathcal{X}_s$ is a disjoint union of $d = \prod_i d_i$ copies of $\operatorname{Spec} \mathbb{C}$, so $\beta^{-1}(s)$ has $d$ points. The semi-continuity of the fibre of $\beta$ implies that $\beta^{-1}(f)$ has $\le d$ points. It follows that $\mathcal{X}_f = \alpha^{-1}(\beta^{-1}(f))$ has $\le d$ connected components.