Solution:

Clearly $f(n)$ is a strictly increasing function, as we can form $f(n-1)$ with $n-1$ ones, and add the final one. However, we can do better; assume Vishal generates $f(n)$ on the board. After $n-2$ minutes, there are two numbers left, say they were formed by $x$ ones and $y$ ones, where $x+y=n$. Clearly the numbers are at most $f(x), f(y)$ (and can be made to be equal to $f(x), f(y)$), and therefore we obtain
$$f(n)=\max _{x+y=n, 1 \leq x \leq y \leq n-1}\left(\max \left(f(x)+f(y), f(x)^{2}\right)\right)$$
where we used the fact that $f$ is increasing to get that $\min \left(f(x)^{2}, f(y)^{2}\right)=f(x)^{2}$ when $x \leq y$. In particular, $f(n+1) \geq f(n)+1$, and $f(2 n) \geq f(n)^{2}$ for all positive integers $n$.

Upper bound:

First proof of upper bound. We use induction. We can check that $f(n)=n$ for $n \leq 4$, and these all satisfy the bound $f(n)=n \leq 3^{n / 3}$. Assume it is true for all $m<n$ (some $n \geq 5$), and with $x, y$ as in equation above we have
$$f(x)^{2} \leq f\left(\left\lfloor\frac{n}{2}\right\rfloor\right)^{2} \leq\left(3^{n / 6}\right)^{2}=3^{n / 3}$$
as desired. It thus remains to show that $f(x)+f(y) \leq 3^{n / 3}$. By induction, it suffices to prove that
$$3^{x / 3}+3^{y / 3} \leq 3^{(x+y) / 3}$$
for $1 \leq x \leq y \leq n-1$ and $x+y=n$. This is equivalent to
$$1+3^{(y-x) / 3} \leq 3^{y / 3}$$
Let $w=3^{(y-x) / 3}$, and we require $3^{x / 3} w \geq w+1$. If $x \geq 2$, then this is true as $w \geq 1$, and if $x=1$ then $w=3^{(n-2) / 3} \geq 3$ and the result is still true. Thus all terms in the equation are at most $3^{n / 3}$, and so $f(n) \leq 3^{n / 3}$, and the upper bound is proven.

Second proof of upper bound. Consider a second game with the same rules but in which Vishal can replace $a$ and $b$ by either $a+b$ or $a b$. Let $g(n)$ be the largest possible value for this new game. Then $f(n) \leq g(n)$ because $\min \{a^{2}, b^{2}\} \leq a b$.

We can check $g(n)=n$ for $n \leq 4$, so $g(n) \leq 3^{n / 3}$ for these values. If $x$ and $y$ are both bigger than $1$, then $g(x)+g(y) \leq g(x) g(y)$. Therefore, for $n>4$, we have that
$$g(n)=\max \{g(n-1)+1, \max _{1 \leq x \leq n-1} g(x) g(n-x)\}$$
Now proceed similarly to the first proof. Assume $n>4$ and $g(m) \leq 3^{m / 3}$ for all $m<n$. If $1 \leq x \leq n-1$, then $g(x) g(n-x) \leq 3^{x / 3} 3^{(n-x) / 3}=3^{n / 3}$. And $g(n-1)+1 \leq 3^{(n-1) / 3}+1$, which is shown to be less than $3^{n / 3}$ in the first proof. It follows that $f(n) \leq g(n) \leq 3^{n / 3}$.

Lower bound:

First proof of lower bound. We begin with a lemma.

Lemma 1. Let $m$ be a nonnegative integer. Then
$$f\left(2^{m}\right) \geq 2^{2^{m-1}} \quad \text{and} \quad f\left(3 \cdot 2^{m}\right) \geq 3^{2^{m}}$$
Proof. We prove the lemma by induction. One can check that $f(n)=n$ for $n \leq 3$, which proves the lemma for $m=0$. For a general $m>0$, we get
$$\begin{gathered} f\left(2^{m}\right) \geq f\left(2^{m-1}\right)^{2} \geq\left(2^{2^{m-2}}\right)^{2}=2^{2^{m-1}} \\ f\left(3 \cdot 2^{m}\right) \geq f\left(3 \cdot 2^{m-1}\right)^{2} \geq\left(3^{2^{m-1}}\right)^{2}=3^{2^{m}} \end{gathered}$$
by induction, as required.

(This lemma can also be proved more constructively. Briefly, if $n=2^{m}$, then partition the $1$'s on the board into $2^{m-1}$ pairs, and then add each pair to get $2^{m-1}$ $2$'s $(2=2^{2^{0}})$; then multiply pairs of $2$'s to get $2^{m-2}$ $4$'s $(4=2^{2^{1}})$; then multiply pairs of $4$'s to get $2^{m-3}$ $16$'s $(16=2^{2^{2}})$; and so on, until there are $2(=2^{1})$ copies of $2^{2^{m-2}}$, which then gets replaced with $2^{2^{m-1}}$. The process is similar for $n=3 \cdot 2^{m}$, except that the first step is to partition the $1$'s into $2^{m}$ groups of $3$, and then use addition within each group to get $2^{m}$ $3$'s on the board.)

Now assume $2^{x} \leq n<3 \cdot 2^{x-1}$ for some integer $x$. Then we have
$$f(n) \geq f\left(2^{x}\right) \geq 2^{2^{x-1}}>2^{n / 3}$$
as required. If no such $x$ exists, then there exists an integer $x$ such that $3 \cdot 2^{x-1} \leq n<2^{x+1}$. In this case, we have
$$f(n) \geq f\left(3 \cdot 2^{x-1}\right) \geq 3^{2^{x-1}}>2^{2^{x+1} / 3}>2^{n / 3}$$
where the second last inequality is equivalent to $2^{x-1} \log (3) \geq \frac{2^{x+1}}{3} \log (2)$, and by dividing out $2^{x}$ and clearing the denominator this is equivalent to $3 \log (3) \geq 4 \log 2$, which is true as $3^{3}=27>16=2^{4}$.

Second proof of lower bound. We shall prove the stronger result $f(n) \geq 2^{(n+1) / 3}$ for $n \geq 2$ by induction. One can check that $f(n)=n$ for $n=2,3,4$, which proves the result for these values. Assume that $n \geq 5$ and that $f(k) \geq 2^{(k+1) / 3}$ for all $k=2,3, \ldots, n-1$. Then
$$\begin{array}{rlr} f(n) & \geq f(\lfloor n / 2\rfloor)^{2} & \\ & \geq\left(2^{(\lfloor n / 2\rfloor+1) / 3}\right)^{2} \quad \quad \text{since }\left\lfloor\frac{n}{2}\right\rfloor \geq 2 \\ & =2^{(2\lfloor n / 2\rfloor+2) / 3} \quad \\ & \geq 2^{(n+1) / 3} \quad \text{since }\left\lfloor\frac{n}{2}\right\rfloor \geq \frac{n-1}{2} . \end{array}$$
The result follows by induction.