Elaborating on the answer by @ajotatxe we can actually show what is the minimum number of steps needed.
As he mentioned, the only sensible moves are $M_k : =\textrm{SC}$k$\textrm{P}$ , that is "select all, copy and then paste $k$ times" where $k \ge 1$.
Each time we apply $M_p$, we multiply the number by $(p+1)$ and use $(p+2)$ steps. Let us call $C(M_p) = p+2$ the cost of the move and $U(M_p) = p+1$ the utility of the move. For a sequence of moves $\bar{M} = (M_{p_1}, \ldots, M_{p_k})$, the cost and utility functions are respectively
$$ C(\bar{M} ) = \sum_{j=1}^k (p_j +2) , \ \ \ U(\bar{M}) = \prod_{j=1}^k (p_j+1) $$
Let us say that a sequence of moves $\bar{M}$ is worse than another sequence $\bar{N}$, denoted $\bar{M} \prec \bar{N}$, if $C(\bar{M}) \ge C(\bar{N})$ and $U(\bar{M}) \le U(\bar{N})$. In practice, we will almost always compare sequences with the same cost. A sequence of moves $\bar{M}$ is called optimal if it is maximal with respect to $\prec$. On a side, note that $\prec$ is only a preorder, that is we can have different sequences with equal utility and cost.
I claim that the only sensible moves are $M_1, M_2, M_3, M_4$. I will show this fact by the following
Lemma. If a sequence of moves $\bar{M}$ contains $M_k$ with $k \ge 5$, there exists $\bar{M} \prec \bar{N} $ with $\bar{N}$ not containing $M_k$ for $k \ge 5$.
Proof. Since composing moves is multiplicative in the utility and additive in the cost, it is enough to show that $M_k$ for $k\ge 5$ is worse than a sequence of moves only made of $M_1, M_2, M_3, M_4$. For $k =5$ we have $M_5 \preceq (M_2, M_1)$ , while for $k \ge 6$ I claim that $M_k \prec (M_{k-5}, M_3)$. Indeed, $k+1 \le 4k - 16 $ whenever $k \ge 17/3 \approx 5.67$.
The second observation is that $M_3$ has the best utility-per-cost; this can be seen by comparing the factor $\sqrt[k+2]{k+1}$ for $k=1,2,3,4$:
$$ \alpha_1 = \sqrt[3]{2} \approx 1.26, \ \ \ \alpha_2 = \sqrt[4]{3} \approx 1.316, \ \ \ \alpha_3 = \sqrt[5]{4} \approx 1.319, \ \ \ \alpha_4 =\sqrt[6]{5} \approx 1.308$$
As a result, we have the following:
Lemma. An optimal sequence contains $M_1$ and $M_4$ at most once and $M_2$ at most four times. Furthermore, $M_2$ and $M_4$ cannot appear together. In particular, all but 5 moves in an optimal sequence are $M_3$ moves.
Proof. Suppose by contradiction that $M_2$ appears five times in an optimal sequence. Note that
$$ C(M_2, M_2, M_2, M_2, M_2) = 20 = C(M_3, M_3, M_3, M_3) $$
The utility of $M_2$ five times is $U(M_2)^5 = \alpha_2^{20}$, while the utility of $M_3$ repeated $4$ times is $\alpha_3^{20}$, concluding the argument.
To prove that $M_4$ and $M_1$ can appear at most once, note that $(M_4, M_4) \prec (M_2, M_2, M_2)$ and $(M_1, M_1) \prec M_4$. The last claim follows from $(M_2, M_4) \prec (M_3, M_3)$.
Lastly, note that $(M_1, M_2, M_2) \prec (M_4, M_3)$ allows for only one $M_2$ when $M_1$ appears. As a result, the only possible optimal, non-$M_3$ moves are the following $7$ moves:
$$M_1, \ \ \ M_1 M_2, \ \ \ M_2, \ \ \ M_2 M_2, \ \ \ M_2 M_2M_2, \ \ \ M_2 M_2 M_2 M_2, \ \ \ M_4 $$
It is possible to show, with the help of a simple program, that these are indeed optimal sequences of moves.
We are ready to show the final theorem:
Theorem. The minimal number of steps $S(n)$ to copy&paste a text $n$ times with select-all, copy and paste moves is given by
$$ S(n) = \min_{i=1, \ldots, 8} 5 \lceil \log_4(n/u_i) \rceil+c_i $$
where $(c_i, u_i)$ varies in the set
$$ I = \{ (0,1), (3,2), (7,6), (4,3), (8,9), (12,27), (16,81), (6,5) \} $$
Proof. The proof is a direct consequence of the above argument. The set $I$ is the set of cost-utility coordinates for the 7 above move sequences, plus an eight element representing the empty move (corresponding to only using $M_3$). If we use a starting sequence with cost-utility $(c,u)$ and then apply $k$ times $M_3$, we get a total cost-utility of $(c+5k, u \cdot 4^k)$. Imposing $u \cdot 4^k \ge n$ we get $k \ge \log_4(n/u)$, so that the smallest integer solution is $k_* = \lceil \log_4(n/u) \rceil$. Substituting into the cost we obtain the claimed formula.
As an application, for $n= 100,000$ we get the best value is obtained with starting move $M_2 M_2 M_2$, and the associated number of steps is... 42, the Answer to the Ultimate Question of Life, The Universe, and Everything!! My compliments to @ajotatxe for having guessed the answer by trial and errors.
To the OP: in hindsight, it's no surprise you and your friend had a hard time finding the right answer. Thanks for the very nice problem!