Answer: $n = 2d + 2$.

Abbreviate Ana, Beto, Carlitos and the candy by $A, B, C$ and $D$ respectively. We also write $d(X, Y)$ for the distance between $X$ and $Y$.

We claim that for $n \le 2d + 1$, $C$ cannot ensure his victory. For the first movement there is no information. Assume without loss of generality that $C$ moves to the left. It could be the case that the initial configuration was the following:

$$\begin{array}{c|c|c|c|c|c} \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \cline{2-6} d & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \cline{1-6} C & \cdots & D & \cdots & A & \cdots \\ \cline{2-6} \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \cline{1-6} \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \cline{1-6} n & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \end{array}$$

Suppose $A$ moves to the left on her first turn. After these moves, $d(C, D) = d + 1$ while $d(A, D) \le n - d - 1 \le d$. Therefore, no matter how $C$ moves, $A$ will get to $D$ before $C$ does if she always moves to the left.

We will now show that for $n = 2d + 2$, $B$ and $C$ can coordinate a strategy to win. We divide the grid into four regions labeled $R_1, R_2, R_3, R_4$ (observe that $C$'s initial cell is not part of any of these regions).

On his first two movements, $C$ moves to the left and to the right, returning to his initial position. This way, after Ana completes her second turn we have $d(A, C) \ge 2d$ and $d(A, D) \ge d$.

Meanwhile, $B$ uses these two turns to tell $C$ what region contains $D$, under the following convention:
$$(\text{hot, hot}) \rightarrow R_1 \quad (\text{hot, cold}) \rightarrow R_2 \quad (\text{cold, hot}) \rightarrow R_3 \quad (\text{cold, cold}) \rightarrow R_4$$

Assume without loss of generality that $D$ is in $R_1$ (for the other cases, analogous strategies are obtained by rotation). $B$ and $C$'s strategy is as follows. In his third turn, $C$ will move upwards. From then on, $B$ will shout “cold” if $C$ is not on the same row as $D$; otherwise he shouts “hot”. In this way, whenever $C$ hears “cold”, he will know that he must move upwards to reduce his distance to $D$. And when he hears “hot”, he starts moving to the right. (Note that it is possible that $C$ will never hear “Hot”, meeting $D$ just by going up.)

Since $d(C, D)$ decreases on each turn, after $d$ turns $C$ gets to $D$. However, to complete the proof we must make sure that $A$ will not get to $C$ or $D$ before $C$ wins the game. Suppose $C$ moved $x < d$ times following the previous strategy. Then $d(A, C) \ge 2d - 2x > 0$ (this is because both $A$ and $C$ move to an adjacent cell on each turn, so their distance is reduced by at most 2), while $d(A, D) \ge d - x > 0$ (because the candy does not move). Hence $A$ cannot reach $C$ or $D$ before $C$ wins the game, as claimed.