Let's look at the single player game which is that I have a budget of $n$ total rolls and I want to come up with a strategy for getting the biggest roll in expectation.
For $k\leq n$, let us look at what happens from $k$ rolls.
Now if we think of each of the rolls $R_1,R_2,\dots,R_k$ as being independent draws from the uniform distribution and we define a new random variable $X:=\max\{R_1,R_2,\dots,R_k\}$, we see that for an arbitrary $x\in[0,1]$, the probability that $X$ is smaller or equal to $x$ is $\mathbb{P}(X\leq x)=\mathbb{P}(R_1\leq x, R_2\leq x,\dots R_k\leq x)=\prod_{j=1}^k \mathbb{P}(R_j\leq x)=\prod_{j=1}^k x=x^k$.
This is clearly the cumulative distribution function of $X$, so we can differentiate that to get the density of $X$, which will obviously be $f_X(x)=kx^{k-1}$. This discussion can also be found with a bit more detail here: https://stats.stackexchange.com/questions/18433/how-do-you-calculate-the-probability-density-function-of-the-maximum-of-a-sample
But if we have the density function, we can immediately compute the expected value of $X$, which will be $\mathbb{E}[X]:=\int_0^1 xf_X(x)dx = \int_0^1 kx^k=\tfrac{k}{k+1}$.
So what does this mean? If I still have $k$ rolls left in my budget, I should expect that the maximum value I will encounter during those remaining $k$ rolls will be $\tfrac{k}{k+1}$.
With this in mind, an optimal strategy for the single player game with a total budget of $n$ rolls is quite straightforward.
Do the first roll and get the value $x_1$. Is $x_1>\tfrac{n-1}{n}$, which is the expected highest value I will see in the remaining $n-1$ rolls? If yes, stop. Otherwise roll again. Get $x_2$ on the second roll. Is $x_2>\tfrac{n-2}{n-1}$? Then stop. Else roll the third time. Inductively, if on roll $\ell$ you have that $x_\ell>\tfrac{n-\ell}{n-\ell+1}$, stop. Otherwise roll for the $\ell+1$-th time. If you are unlucky enough to get to the $n$-th roll, the value you will stop at will be arbitrary.
But now, what is the expected outcome of the optimal strategy? Note that you would stop at roll $1$ with probability $S_1=\tfrac{1}{n}$ and you will only do that when $x_1\in(\tfrac{n-1}{n},1]$. The average value of stopping at step $1$ will clearly be $A_1=\tfrac{1}{2}\big(\tfrac{n-1}{n}+1\big)=\tfrac{2n-1}{2n}$.
You stop at roll $2$ if you did not stop at roll $1$ and if $x_2\in(\tfrac{n-2}{n-1},1]$. The probability that you did not stop at roll $1$ is $1-\tfrac{1}{n}=\tfrac{n-1}{n}$ and the probability that you roll $x_2>\tfrac{n-2}{n-1}$ is $\tfrac{1}{n-1}$. So the probability that you stop at roll $2$ is $S_2=\tfrac{n-1}{n}\tfrac{1}{n-1}=\tfrac{1}{n}$. The average stopping value on roll $2$ is $A_2=\tfrac{2n-3}{2n-2}$.
You stop at roll $3$ if you did not stop at roll $1$, did not stop at roll $2$ and if $x_3\in(\tfrac{n-3}{n-2},1]$. This will happen with probability $S_3=(1-2\tfrac{1}{n})\tfrac{1}{n-2}=\tfrac{1}{n}$ and your average stopping value will be $A_3=\tfrac{2n-5}{2n-4}$.
You can show by induction that $S_k=\tfrac{1}{n}$ since you stop at roll $k$ if you have not stopped at any of the previous rolls (which happens with probability $1-\sum_{j=1}^{k-1} S_j=\tfrac{n-k}{n}$ by the induction hypothesis) and if $x_k>\tfrac{n-k-1}{n-k}$ which has probability $\tfrac{1}{n-k}$. The average outcome will be $A_k=\tfrac{2n-2k+1}{2n-2k+2}$. This will hold for all $2\leq k\leq n-1$. Stopping at roll $n$ will occur with probability $S_n=1-\sum_{j=1}^{n-1} S_j=\tfrac{1}{n}$ with average outcome $A_n=\tfrac{1}{2}=\tfrac{2n-2n+1}{2n-2n+2}$.
The expected value of the single player strategy is then $E_n=\sum_{k=1}^n S_kA_k= \tfrac{1}{n} \sum_{k=1}^n \tfrac{2n-2k+1}{2n-2k+2} =\tfrac{1}{n} \sum_{k=1}^n\big(1-\tfrac{1}{2k}\big)=1-\tfrac{1}{2n}\sum_{k=1}^n \tfrac{1}{k}$.
$E_2=1-\tfrac{1}{4}\big(1+\tfrac{1}{2}\big)=\tfrac{5}{8}$.
$E_3=1-\tfrac{1}{6}\big(1+\tfrac{1}{2}+\tfrac{1}{3}\big)=\tfrac{25}{36}$.
As expected $E_3>E_2$. I am unsure what the game theory perspective on this is though. If both players follow the single-player strategy, the player with $3$ rolls is definitely expected to win. How the other player would want to adapt his strategy depends on some factors. For instance, do both players know what roll the other player is on or whether they stopped early?
Major edit thanks to @hgmath
What I described above is not the optimal single player strategy. Indeed, consider the case when $n=3$. If on roll one we get $x_1\in(E_2,\tfrac{2}{3})=(\tfrac{5}{8},\tfrac{2}{3})$, rolling again and pursuing the above strategy would be a mistake, since our expected outcome would just be $E_2$ (the upper bound $\tfrac{2}{3}$ is put there since even with the current strategy we would not reroll if we got above $\tfrac{2}{3}$).
This suggests an inductive strategy. Namely, if we already know the optimal expected outcome of $E_{n-1}$ rolls, we should only reroll after the first roll if $x_1<E_{n-1}$ instead of rerolling if $x_1<\tfrac{n-1}{n}$. So let us try to write down these optimal expectations $E_k$.
If $n=1$, i.e., if our roll budget is $1$, clearly we will stop after $x_1$ with probability $S_1=1$ and the average outcome will be $A_1=\tfrac{1}{2}$, meaning that the optimal $E_1=S_1A_1=\tfrac{1}{2}$.
If our roll budget is $n=2$, then after the first roll $x_1$, we need to check if $x_1>E_1=\tfrac{1}{2}$. This will happen with probability $S_1=1-E_1=\tfrac{1}{2}$ and the average outcome will be $A_1=\tfrac{1+E_1}{2}=\tfrac{3}{4}$. If $x_1\leq E_1$, which will happen with probability $E_1=\tfrac{1}{2}$, we stop at $S_2=1-S_1=\tfrac{1}{2}$ with average outcome $A_2=\tfrac{1}{2}$. Thus the expected optimal outcome is $E_2=S_1A_1+S_2A_2=\tfrac{1}{2}\big((1-E_1)(1+E_1)+E_1\big)=\tfrac{1}{2}(1+E_1-E_1^2)=\tfrac{5}{8}$. So far no difference to our previous computation.
However, if $n=3$, we should stop at $x_1$ if $x_1>E_2$, with probability $S_1=1-E_2=\tfrac{3}{8}$ and average outcome $A_1=\tfrac{1+E_2}{2}=\tfrac{13}{16}$. If $x_1\leq E_2$, then we roll the second time to get $x_2$. We stop if $x_2>E_1$, an event with probability $S_2=(1-S_1)(1-E_1)=E_2(1-E_1)=\tfrac{5}{8}\tfrac{1}{2}=\tfrac{5}{16}$, and an average outcome $A_2=\tfrac{1+E_1}{2}=\tfrac{3}{4}$. If $x_2<E_1$, we roll again and we are forced to stop with $x_3$, an event with probability $S_3=1-S_1-S_2=E_2-E_2(1-E_1)=E_1E_2=\tfrac{5}{16}$ and average outcome $A_3=\tfrac{1}{2}$.
Thus, the expected outcome of the optimal $3$ roll strategy is $E_3=S_1 A_1+S_2A_2+S_3A_3=\tfrac{1}{2}\big((1-E_2)(1+E_2)+E_2(1-E_1)(1+E_1)+E_1E_2\big) = \tfrac{1}{2}(1+E_2+E_2E_1-E_2^2-E_2E_1^2)=\tfrac{1}{2}\big(1+\tfrac{5}{8}+\tfrac{5}{16}-\tfrac{25}{64}-\tfrac{5}{32}\big)=\tfrac{89}{128}$.
This is bigger than what we got with the old strategy by an eight of a percent.
If $n=4$, we would get $S_1=1-E_3$ with $2\cdot A_1=1+E_3$, $S_2=(1-S_1)(1-E_2)=E_3(1-E_2)$ with $2\cdot A_2=1+E_2$, $S_3=(1-S_1-S_2)(1-E_1)=E_3E_2(1-E_1)$ with $2\cdot A_3=1+E_1$ and $S_4=(1-S_1-S_2-S_3)=E_3E_2E_1$ with $2\cdot A_4=1$, meaning that $E_4=\tfrac{1}{2}(1+E_3+E_3E_2+E_3E_2E_1-E_3^2-E_3E_2^2-E_3E_2E_1^2)$. After some computation, this means $E_4=\tfrac{25195}{32768}$. This is an almost $4\%$ increase over the expected outcome of our previous strategy, which would have been $1-\tfrac{1}{8}(1+\tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{4})=\tfrac{71}{96}$.
It is not difficult to find a recursive relation for $E_n$ based on induction and the obvious pattern from the computed cases, but I don't know if there is any reasonable way to get a closed form expression for $E_n$.