Students who see ears of another student

Solution 1:

The minimum is $n+2$, with the construction in chubakueno's comment.

Proof that this is minimal:

Split the students into lines (maximal subsets of students in a straight line all looking the same direction).

The last student in each line must be looking at an ear (in fact, the number of lines equals the number of students looking at an ear).

Each line has at most $n-1$ students.

Thus there are at least $\lceil \frac{n^2}{n-1} \rceil = n+2$ lines, and thus at least this many students looking at an ear.

An illustration (slight modification of chubakueno's comment):

$$\begin{matrix} \downarrow&\downarrow&\downarrow&\downarrow&\leftarrow\\ \downarrow&\downarrow&\downarrow&\downarrow&\uparrow\\ \downarrow&\downarrow&\downarrow&\downarrow&\uparrow\\ \downarrow&\downarrow&\downarrow&\downarrow&\uparrow\\ \rightarrow&\rightarrow&\rightarrow&\rightarrow&\uparrow \end{matrix}$$

You see $n+1$ lines each with the maximum $n-1$ elements, plus one lonely student in a line of one element (upper right).

Solution 2:

Consider a slight modification of your arrangement: $$\begin{matrix} \downarrow&\downarrow&\downarrow&\downarrow&\downarrow\\ \downarrow&\downarrow&\downarrow&\downarrow&\downarrow\\ \downarrow&\downarrow&\downarrow&\downarrow&\color{red}\downarrow\\ \color{red}\downarrow&\color{red}\downarrow&\color{red}\downarrow&\color{red}\downarrow&\color{red}\leftarrow\\ \rightarrow&\rightarrow&\rightarrow&\color{red}\rightarrow&\color{red}\uparrow \end{matrix}$$ which looks better to me, and in the general case, gives $n+3$ as the number of students who see the side of another student's face for $n \ge 3$, and for $n = 2$, the minimum is obviously $4$.