Solution:

We will show that the maximum number of good cells over all possible starting configurations is
$$\frac{n^{2}}{4} \quad \text{if } n \text{ is even and}$$
$$0 \quad \text{if } n \text{ is odd.}$$

## Odd $n$

First, we will prove that there are no good cells if $n$ is an odd number.
For Turbo to reach her goal, she must return to her initial cell after visiting every cell exactly once. Consider the chessboard coloring of the board. Without loss of generality, we assume that Turbo starts in a black cell. Since, at every step, Turbo moves to a cell of a different color; she will be in a white cell after $n^{2} \equiv 1 \mod 2$ moves. Thus, it is impossible for Turbo to come back to her initial black cell on her $n^{2}$-th move, which is a contradiction. Thus there are no good cells.

## Lower bound for even $n$

We will now construct a starting configuration with $\frac{n^{2}}{4}$ good cells for even $n$.
Let $(i,j)$ denote the cell in row $i$ and column $j$. Consider the following cycle
$$\begin{array}{r l} & (1,1)\rightarrow (1,2)\rightarrow (1,3)\rightarrow \ldots \rightarrow (1,n)\\ & \qquad \rightarrow (2,n)\rightarrow (2,n - 1)\rightarrow \ldots \rightarrow (2,2)\\ & \qquad \ldots\\ & \qquad \rightarrow (2i - 1,2)\rightarrow (2i - 1,3)\rightarrow \ldots \rightarrow (2i - 1,n)\\ & \qquad \rightarrow (2i,n)\rightarrow (2i,n - 1)\rightarrow \ldots \rightarrow (2i,2)\\ & \qquad \ldots\\ & \qquad \rightarrow (n,n)\rightarrow (n,n - 1)\rightarrow \ldots \rightarrow (n,2)\\ & \qquad \rightarrow (n,1)\rightarrow (n - 1,1)\rightarrow \ldots \rightarrow (2,1)\rightarrow (1,1). \end{array} \quad (1,1)$$

Note that the cycle returns to the initial cell after visiting every cell exactly once. To prove that $(1,1)$ is good, we need to find a starting configuration such that Turbo traverses this cycle.
Let $c_{i}$ be the $(i - 1)$-th cell on the cycle: so we have $c_{0} = (1,1)$, $c_{2} = (1,2)$, ..., $c_{n^{2} - 1} = (2,1)$. For every $i$, we draw an arrow in cell $c_{i}$ pointing towards cell $c_{i + 1}$ (or pointing towards $c_{0}$ if $i = n^{2} - 1$) and then rotate this arrow $i$ times $90^{\circ}$ in the clockwise direction. After $i$ moves, the arrow in $c_{i}$ will have rotated $i$ times $90^{\circ}$ counterclockwise and be in the same direction as on the path defined above. Thus, Turbo will traverse the cycle $c_{0}, c_{1}, c_{2}, \ldots , c_{n^{2} - 1}, c_{0}$ and $(1,1)$ is good.
Every four moves, all arrows point in the same direction as in the beginning. Moreover, the board will return to its initial configuration after traversing the full cycle, since $n^{2}$, the length of the cycle, is divisible by 4. Therefore Turbo can also start at any $c_{i}$ with $4 \mid i$ and follow the same route. Hence the cells $c_{0}, c_{4}, c_{8}, \ldots , c_{n^{2} - 4}$ are good and there are $\frac{n^{2}}{4}$ of such cells.

## Upper bound for even $n$

We will prove that for even $n$ and any start configuration there are at most $\frac{n^{2}}{4}$ good cells.
Let $a_{0}$ be a good cell. Let $a_{0},a_{1},a_{2},\ldots ,a_{n^{2} - 1},a_{n^{2}} = a_{0}$ be the sequence of cells that Turbo visits when she starts at $a_{0}$. Now suppose there is another good cell $b_{0}$ and let $b_{0},b_{1},b_{2},\ldots ,b_{n^{2} - 1},b_{n^{2}} = b_{0}$ be the sequence of cells that Turbo visits when she starts at $b_{0}$.
Note that, since $4\mid n^{2}$, the arrows are back to their initial configuration after $n^{2}$ steps. Thus, if Turbo keeps walking after returning to her initial cell, she would just traverse the same cycle over and over again.
Consider the upper left corner of the board. With standard row and column numbering, the corner cell is $(1,1)$. This cell has only two neighbours, so both the $a$-route and the $b$-route must have cells $(2,1),(1,1),(1,2)$ in that order or $(1,2),(1,1),(2,1)$ in that order. Without loss of generality, $a_{i - 1} = (2,1)$, $a_{i} = (1,1)$ and $a_{i + 1} = (1,2)$ for some $i$. Let $j$ be such that $b_{j} = (1,1)$. If $b_{j - 1} = (2,1) = a_{i - 1}$ then the arrow in cell $(2,1)$ must be pointed in the same direction after $i - 1$ steps and after $j - 1$ steps, so $i\equiv j \pmod{4}$. But then the arrow in cell $b_{j} = (1,1) = a_{i}$ must also be pointed in the same direction after $i$ and after $j$ steps, so Turbo moves to $b_{j + 1} = a_{i + 1}$ in both cases, and again finds the arrow pointed in the same direction in both cases. Continuing, we find that the $b$-route is actually identical to $a_{4t},a_{4t + 1},\ldots ,a_{n^{2}} = a_{0},a_{1},\ldots ,a_{4t - 1},a_{4t}$ for some $t$, as any other starting point would have the arrows in the wrong direction initially.

Now suppose instead that $b_{j + 1} = (2,1) = a_{i - 1}$. Considering the $a$-route, the arrows in the upper left corner after $i - 1$ steps must be like this:

Considering the $b$-route instead, the arrows after $j - 1$ steps must be like this:

From the arrows in cell $(1,1)$ we see that $i\equiv j + 1 \pmod{4}$. However, for the cells $(2,1)$ and $(1,2)$ this gives a contradiction.
We conclude that the only possible good cells are $a_{4t}$ for $t = 0,1,\ldots ,\frac{n^{2}}{4} - 1$, which gives at most $\frac{n^{2}}{4}$ good cells.