Let $A$ be a finite set of positive integers and $\pi$ be a permutation (=bijection) on $A$. A pair $(x, \pi(x)) \in A^2$ is called a "up" if $x < \pi(x)$ holds, and $up(\pi)$ is defined by the number of up's in $\pi$.

Given a permutation $\pi$ on $A = \{x_1 < x_2 < \dots < x_m\}$, define a permutation $\pi'$ by
$$\pi' = x_2x_3\cdots x_{1} \circ \pi,$$
where $\circ$ means the functional composition. Then we can easily deduce the following lemma.

Lemma. Let $\pi$ be a permutation on $A = \{x_1 < x_2 < \dots < x_m\}$ without fixed points. Then we have
$$up(\pi') = up(\pi) - 1.$$

Let $S_0 = \{s_1, s_2, \dots, s_n\}$ be the set of $n$ students and $T_0 = \{t_1, t_2, \dots, t_n\}$ be the set of $n$ name tags. A distribution of name tags can be interpreted as a bijection $\tau_0: T_0 \to S_0$. By simplifying $\tau_0(t_i) = s_j$ to $\tau_0(i) = j$, we can regard $\tau$ as a permutation on $n$. Let $F_1$ be the set of fixed points of $\tau_0$. Define $S_1 = S_0 - \{s_i \mid i \in F_1\}$ and $T_1 = T_0 - \{t_i \mid i \in F_1\}$. Then $\tau_1: T_1 \to S_1$ represents the state after the exit of each student having one's own name tag. It is obvious that $up(\tau_0) = up(\tau_1)$. Also we can easily deduce the following:

Given a distribution $\pi$, passing their name tags to the right derives a distribution $\pi'$.
Therefore $\tau'_1$ is the distribution after one process. By the lemma we have
$$up(\tau'_1) = up(\tau_1) - 1 = up(\tau_0) - 1.$$

Thus, with a single process, the number of up's is decreased by 1. Thus if an initial distribution $\tau_0$ has at most 3 up's then every student has one's own name tag during 3 processes, i.e., every student gets out of the circle until 4 processes. Therefore the answer of this problem is
$$n! - |\{\tau_0 \in \text{Sym}_n \mid \text{up}(\tau_0) \le 3\}|,$$
where $\text{Sym}_n$ is the set of permutations on $\{1, 2, \dots, n\}$.

Let $B(n, k)$ be the set of $\{\pi \in \text{Sym}_n \mid \text{up}(\pi) \le k\}$ and $b(n, k) := |B(n, k)|$. By considering the position of $n$ in $\pi \in B(n, k)$, we can easily deduce the following recurrence relation:
$$b(n, k) = (k + 1)b(n - 1, k) + (n - k - 1)b(n - 1, k - 1), \quad n \ge 2$$
with $b(n, 0) = 1$ and $b(n, k) = 0$ for $k \ge n$. For $k = 1$ we have
$$b(n, 1) = 2b(n - 1, 1) + (n - 2). \quad b(1, 1) = 1.$$
The recurrence relation yields $b(n, 1) = 2^n - n$. For $k = 2$ we have
$$b(n, 2) = 3b(n - 1, 2) + (n - 3)(2^{n-1} - n + 1), \quad b(1, 2) = 1.$$
The recurrence relation yields $b(n, 2) = 3^n - n2^n + \binom{n}{2}$. Finally, for $k = 3$, we have
$$b(n, 3) = 4b(n - 1, 3) + (n - 4) \left( 3^{n-1} - (n-1)2^{n-1} + \binom{n-1}{2} \right), \quad b(1, 3) = 1.$$
The recurrence relation yields $b(n, 3) = 4^n - n3^n + \binom{n}{2}2^n - \binom{n}{3}$. Therefore the answer of this problem is
$$n! - b(n, 3) = n! - 4^n + n3^n - \binom{n}{2}2^n + \binom{n}{3}.$$