$\def\i{\mathrm{i}}$As is pointed out by @PaoloLeonetti, for even $n$, the minimum is $0$. Now suppose $n = 2m + 1$ where $m \in \mathbb{N}$. It will be proved that$$
\sum_{k = 1}^{2m + 1} |z_k^2 + 1| \geqslant 1.
$$
The equality can be achieved for $z_1 = 0,\ z_k = (-1)^k \i\ (k \geqslant 2)$.
Lemma: If $a_1, \cdots, a_{2m + 1} \in \mathbb{R}$, $|a_k| \leqslant 1$ for $1 \leqslant k \leqslant 2m + 1$, and $\sum\limits_{k = 1}^{2m + 1} a_k = 0$, then$$
\sum_{k = 1}^{2m + 1} a_k^2 \leqslant 2m.
$$
Proof of lemma: If there exists $k_0$ such that $a_{k_0} = 0$, then$$
\sum_{k = 1}^{2m + 1} a_k^2 = \sum_{k \neq k_0} a_k^2 \leqslant \sum_{k \neq k_0} 1^2 = 2m.
$$
Now suppose $a_k \neq 0$ for each $k$. Since there are $2m + 1$ nonzero numbers, then either there are at least $m + 1$ positive numbers or there are at least $m + 1$ negative numbers. Without loss of generality, suppose $a_1, \cdots, a_k > 0$, $k \geqslant m + 1$, and $a_1 = \min\limits_{1 \leqslant j \leqslant k} a_k$. Note that$$
\sum_{j = 1}^k a_j = -\sum_{j = k + 1}^{2m + 1} a_j \leqslant 2m + 1 - k \leqslant m. \tag{1}
$$
Denote $S = \sum\limits_{j = 1}^{2m + 1} a_j^2$. Take$$a_{1,1} = a_1 - \min(a_1, 1 - a_2),\ a_{1,2} = a_2 + \min(a_1, 1 - a_2),\ a_{1,j} = a_j\ (j \geqslant 3),$$
then\begin{align*}
S_1 - S &= \sum_{j = 1}^{2m + 1} a_{1,j}^2 - \sum_{j = 1}^{2m + 1} a_j^2 = (a_{1,1}^2 + a_{1,2}^2) - (a_1^2 + a_2^2)\\
&= 2(a_2 - a_1) \min(a_1, 1 - a_2) + (\min(a_1, 1 - a_2))^2 \geqslant 0,
\end{align*}
and$$
\sum_{j = 1}^k a_{1,j} = \sum_{j = 1}^k a_j, \quad |a_{1,j}| \leqslant 1\ (1 \leqslant j \leqslant 2m + 1), \quad a_{1,1} = \min_{1 \leqslant j \leqslant k} a_{1, j}.$$
If $a_{1,j} = 0$, then again$$
\sum_{j = 1}^{2m + 1} a_j^2 \leqslant \sum_{j = 1}^{2m + 1} a_{1,j}^2 = \sum_{j > 1} a_{1,j}^2 \leqslant \sum_{j > 1} 1^2 = 2m.
$$
Otherwise $a_{1,2} = 1$. Take$$
a_{2,1} = a_{1,1} - \min(a_{1,1}, 1 - a_{1,3}),\ a_{2,3} = a_{1,3} + \min(a_{1,1}, 1 - a_{1,3}),\ a_{2,j} = a_{1,j}\ (j \neq 1, 3),
$$
then analogously there is $S_2 \geqslant S_1$ with$$
\sum_{j = 1}^k a_{2,j} = \sum_{j = 1}^k a_{1,j}, \quad |a_{2,j}| \leqslant 1\ (1 \leqslant j \leqslant 2m + 1), \quad a_{2,1} = \min_{1 \leqslant j \leqslant k} a_{2, j}.$$
Now it will be proved that such adjustments can be carried out for at most $k - 1$ times, i.e. there exists $1 \leqslant t \leqslant k - 1$ such that$$
a_1 \geqslant a_{1,1} \geqslant \cdots \geqslant a_{t,1} = 0.
$$
If otherwise, then $a_{k - 1, 1} > 0$ and $a_{k - 1, j} = 1$ for $2 \leqslant j \leqslant k$. Thus$$
\sum_{j = 1}^k a_j = \sum_{j = 1}^k a_{k - 1, j} > \sum_{j = 2}^k a_{k - 1, j} \geqslant k - 1 \geqslant m,
$$
contradictory to (1). Therefore, suppose $1 \leqslant t \leqslant k - 1$ satisfies$$
a_1 \geqslant a_{1,1} \geqslant \cdots \geqslant a_{t,1} = 0,
$$
then$$
\sum_{j = 1}^{2m + 1} a_j^2 \leqslant \sum_{j = 1}^{2m + 1} a_{t,j}^2 = \sum_{j > 1} a_{t,j}^2 \leqslant \sum_{j > 1} 1^2 = 2m.
$$
Now back to the question. Suppose $z_k = x_k + \i y_k \ (x_k, y_k \in \mathbb{R})$ for each $k$. If there exists $k$ such that $x_k \neq 0$, without loss of generality suppose $k = 1$ and $x_1 > 0$, then by $\sum z_k = 0$, there exists $l \neq 1$ such that $x_l < 0$. Without loss of generality suppose $l = 2$.
Take$$
x_1' = x_1 - \min(x_1, -x_2) \geqslant 0,\ x_2' = x_2 + \min(x_1, -x_2) \leqslant 0,\ x_k' = x_k \ (k \geqslant 3),
$$
and $z_k' = x_k' + \i y_k$ for each $k$, then$$
|z_1 + \i|^2 - |z_1' + \i|^2 = x_1^2 - x_1'^2 > 0 \Longrightarrow |z_1 + \i| > |z_1' + \i|,\\
|z_2 + \i|^2 - |z_2' + \i|^2 = x_2^2 - x_2'^2 > 0 \Longrightarrow |z_2 + \i| > |z_2' + \i|.
$$
Analogously,$$
|z_1 - \i| > |z_1' -\i|, \quad |z_2 - \i| > |z_2' -\i|.
$$
Therefore,\begin{align*}
&\mathrel{\phantom{=}} \sum_{k = 1}^{2m + 1} |z_k^2 + 1| - \sum_{k = 1}^{2m + 1} |z_k'^2 + 1| = (|z_1^2 + 1| + |z_2^2 + 1|) - (|z_1'^2 + 1| + |z_2'^2 + 1|)\\
&= (|z_1 + \i| \cdot |z_1 - \i| - |z_1' + \i| \cdot |z_1' - \i| ) + (|z_2 + \i| \cdot |z_2 - \i| - |z_2' + \i| \cdot |z_2' - \i| ) > 0.
\end{align*}
This implies if there exists $k$ such that $x_k \neq 0$, then there is another tuple $(z_1', \cdots, z_{2m + 1}')$ satisfies $\sum z_k' = 0$ and $\sum z_k'^2 < \sum z_k^2$. Therefore,$$
\min_{\substack{z_1, \cdots, z_n \in \mathbb{C}\\\sum z_k = 0}} \sum_{k = 1}^{2m + 1} |z_k^2 + 1| = \min_{\substack{z_1, \cdots, z_n \in \mathbb{C}\\\sum z_k = 0\\\mathrm{Re}(z_k) = 0\ (1 \leqslant k \leqslant 2m + 1)}} \sum_{k = 1}^{2m + 1} |z_k^2 + 1|.
$$
Now it suffices to prove that for any $y_1, \cdots, y_{2m + 1} \in \mathbb{R}$, if $\sum\limits_{k = 1}^{2m + 1} y_k = 0$, then$$
\sum_{k = 1}^{2m + 1} |1 - y_k^2| \geqslant 1.
$$
If $M = \max\limits_{1 \leqslant k \leqslant 2m + 1} |y_k| > 1$, without loss of generality, suppose $y_1 = M$, then there exists $k$ such that $y_k < 0$. Without loss of generality, suppose $y_2 < 0$. Take$$
y_1' = 1,\ y_2' = y_2 + y_1 - 1, y_k' = y_k\ (k \geqslant 3),
$$
then\begin{align*}
&\mathrel{\phantom{=}} \sum_{k = 1}^{2m + 1} |1 - y_k^2| - \sum_{k = 1}^{2m + 1} |1 - y_k'^2| = |1 - y_1^2| + |1 - y_2^2| - |1 - y_2'^2|\\
&= (y_1^2 - 1) + |1 - y_2^2| - |(y_1 + y_2)(y_1 + y_2 - 2)|\\
&= (y_1^2 - 1) + |1 - y_2^2| - (y_1 + y_2)|y_1 + y_2 - 2|, \tag{2}
\end{align*}
where the last identity is due to $y_1 = |y_1| \geqslant |y_2| = -y_2$, i.e. $y_1 + y_2 \geqslant 0$.
Case 1: $y_1 + y_2 \geqslant 2$. Then by $y_1 > 1$ and $y_2 < 0$, there is\begin{align*}
(2) &= (y_1^2 - 1) + |1 - y_2^2| - (y_1 + y_2)(y_1 + y_2 - 2)\\
&\geqslant (y_1^2 - 1) + (y_2^2 - 1) - (y_1 + y_2)(y_1 + y_2 - 2)\\
&= -2(y_1 - 1)(y_2 - 1) > 0.
\end{align*}
Case 2: $0 < y_1 + y_2 < 2$. Then\begin{align*}
(2) &= (y_1^2 - 1) + |1 - y_2^2| + (y_1 + y_2)(y_1 + y_2 - 2)\\
&\geqslant (y_1^2 - 1) + (1 - y_2^2) + (y_1 + y_2)(y_1 + y_2 - 2)\\
&= 2(y_1 - 1)(y_1 + y_2) > 0.
\end{align*}
Case 3: $y_1 + y_2 = 0$. Then by $y_2 = -y_1 < -1$, there is$$
(2) = (y_1^2 - 1) + (y_2^2 - 1) - 0 = 2(y_1^2 - 1) > 0.
$$
Threefore,$$
\sum_{k = 1}^{2m + 1} |1 - y_k'^2| < \sum_{k = 1}^{2m + 1} |1 - y_k^2|.
$$
This implies$$
\min_{\substack{y_1, \cdots, y_{2m + 1} \in \mathbb{R}\\\sum y_k = 0}} \sum_{k = 1}^{2m + 1} |1 - y_k^2| = \min_{\substack{y_1, \cdots, y_{2m + 1} \in \mathbb{R}\\\sum y_k = 0\\|y_k| \leqslant 1\ (1 \leqslant k \leqslant 2m + 1)}} \sum_{k = 1}^{2m + 1} |1 - y_k^2|.
$$
Now it suffices to prove that for any $y_1, \cdots, y_{2m + 1} \in \mathbb{R}$, if $|y_k| \leqslant 1$ for each $k$ and $\sum\limits_{k = 1}^{2m + 1} y_k = 0$, then$$
\sum_{k = 1}^{2m + 1} (1 - y_k^2) \geqslant 1,
$$
i.e. $\sum y_k^2 \leqslant 2m$, which is true by the lemma.
A simpler proof of lemma: Without loss of generality, suppose $a_1, \cdots, a_k \geqslant 0$ and $a_{k + 1}, \cdots, a_{2m + 1} \leqslant 0$.
If $\sum\limits_{j = 1}^k a_j^2 \leqslant k - 1$, then$$
\sum_{j = 1}^{2m + 1} a_j^2 = \sum_{j = 1}^k a_j^2 + \sum_{j = k + 1}^{2m + 1} a_j^2 \leqslant (k - 1) + (2m - k + 1) = 2m.
$$
If $\sum\limits_{j = k + 1}^{2m + 1} a_j^2 \leqslant 2m - k$, analogously there is $\sum\limits_{j = 1}^{2m + 1} a_j^2 \leqslant 2m$.
Now suppose$$
\sum_{j = 1}^k a_j^2 > k - 1, \quad \sum_{j = k + 1}^{2m + 1} a_j^2 > 2m - k. \tag{3}
$$
Because$$
k - 1 < \sum_{j = 1}^k a_j^2 \leqslant \sum_{j = 1}^k a_j = -\left( \sum_{j = k + 1}^{2m + 1} a_j \right) \leqslant 2m - k,
$$
then $2k < 2m + 1$, which implies $k \leqslant m$. Thus$$
m \leqslant 2m - k < \sum_{j = k + 1}^{2m + 1} a_j^2 \leqslant -\left( \sum_{j = k + 1}^{2m + 1} a_j \right) = \sum_{j = 1}^k a_j \leqslant k \leqslant m,
$$
a contradiction. Therefore, (3) cannot hold.