Find the generators of multiplicative group of units efficiently?
For any $g$ in the set $\mathbb Z_p^*=\{1,2,\dots,p-1\}$, consider the function $F_g$ over that set defined by $F_g(x)\;=\;g\cdot x\bmod p$. Since $p$ is prime, by Fermat's little theorem, iterating $F_g$, starting from $1$, $p-1$ times, cycles back to $1$.

By definition, $g$ is a generator of $\mathbb Z_p^*$ if and only if that cycling does not occur before these $p-1$ iterations. The smallest $k&gt;0$ where the cycling occurs divides $p-1$ (that $k$ is the order of $g$). Proof sketch: if that $k$ did not divide $p-1$, $k'=(p-1)\bmod k$ would be a smaller $k'&gt;0$ for which the cycling occurs.

It follows that we can test if some $g$ not divisible by $p$ is a generator of $\mathbb Z_p^*$ by checking if $g^k\bmod p\ne1$, with $k=(p-1)/q$ for $q$ each of the distinct prime factors of $p-1$.

Here, for $g$ each of the integers in list $L$, we check if $g\bmod p\ne0$, and $g^{(p-1)/2}\bmod p\ne1$, and $g^{(p-1)/p_1}\bmod p\ne1$, and $g^{(p-1)/p_2}\bmod p\ne1$, and $g^{(p-1)/p_3}\bmod p\ne1$. In the affirmative, and only then, $g$ is a generator of $\mathbb Z_p^*$.

Additions:


Computing $g^k\bmod p$ is feasible with a modest computer, and common, for integers of many thousands of bits: $2\log_2(k)$ modular multiplications are enough to compute $g^k\bmod p$; see modular exponentiation.
If the list $L$ was not a given, we could try $g$ at random (or small primes by increasing orders, which is slightly more efficient than trying small integers by increasing order). If all the primes dividing $(p-1)/2$ are large (which is the case here), nearly 50% of candidates will work, thus a search won't be too long.
Often, we want a generator of a subgroup of order one of the large primes dividing $p-1$, say $p_3$; we can get that as $g'=g^{(p-1)/p_3}\bmod p$.