Let's try it!

To make finding $g$ easier, let's choose a prime $p$ such that $(p-1)/2=q$ is prime (so $p-1 = 2q$, with $q$ prime). Since for any $g$ with $\gcd(g,p)=1$,

$$g^{2q} \equiv 1\pmod{p},$$

the order of $g$ is $1$, $2$, $q$, or $2q=p-1$, so the order of $g$ is easy to compute.

For our first example, let $p=23$. Then $g=5$ has order $p-1=22$. (I found $g=5$ using the function znprimroot in PARI. You can also just compute the order of $2$, $3$, etc., until you find a number with order $p-1$.)

Nikita: Chooses secret $n=12$; sends $g^{12} = 5^{12} \equiv \mathbf{18} \pmod{23}$.

Michael: Chooses secret $n=5$; sends $g^5 = 5^5 \equiv \mathbf{20} \pmod{23}$.

Compute Shared Secret:
Nikita: $20^{12} \equiv \mathbf{3}\pmod{23}$
Michael: $18^{5} \equiv \mathbf{3}\pmod{23}.$