There seems to be multiple interpretations of this question.
Every broken diagonal contains every symbol
Possible if and only if $n$ is odd.
We can use diagonally cyclic Latin squares (as in DrLemniscate's answer). For all odd $n \geq 1$, we use this particular square:
$$
\begin{bmatrix}
\color{blue} 8 & 7 & 6 & 5 & 4 & \color{red} 3 & 2 & 1 & 0 \\
1 & \color{blue}0 & 8 & 7 & 6 & 5 & \color{red} 4 & 3 & 2 \\
3 & 2 & \color{blue} 1 & 0 & 8 & 7 & 6 & \color{red} 5 & 4 \\
5 & 4 & 3 & \color{blue} 2 & 1 & 0 & 8 & 7 & \color{red} 6 \\
\color{red} 7 & 6 & 5 & 4 & \color{blue} 3 & 2 & 1 & 0 & 8 \\
0 & \color{red} 8 & 7 & 6 & 5 & \color{blue} 4 & 3 & 2 & 1 \\
2 & 1 & \color{red} 0 & 8 & 7 & 6 & \color{blue} 5 & 4 & 3 \\
4 & 3 & 2 & \color{red} 1 & 0 & 8 & 7 & \color{blue} 6 & 5 \\
6 & 5 & 4 & 3 & \color{red} 2 & 1 & 0 & 8 & \color{blue} 7 \\
\end{bmatrix}
$$
Horizontally the symbols decrease by $1$; vertically the symbols increase by $2$; along the main broken diagonals (and in particular, along the main diagonal) the symbols increase by $1$. These properties ensure that there are no repeated symbols in rows and columns (i.e., it is a Latin square), and there are no repeated symbols on the broken diagonals. (Formally, $-1$, $1$, and $2$ are generators of $\mathbb{Z}_n$ when $n$ is odd.)
Now at this point we notice that diagonally cyclic Latin squares have an orthogonal mate where each main diagonal is a unique symbol:
$$
\begin{bmatrix}
\color{blue} 2 & \color{red} 1 & \bf \color{green} 0 \\
\bf \color{green} 1 & \color{blue} 0 & \color{red} 2 \\
\color{red} 0 & \bf \color{green} 2 & \color{blue} 1 \\
\end{bmatrix}
\begin{bmatrix}
\color{blue} 0 & \color{red} 1 & \bf \color{green} 2 \\
\bf \color{green} 2 & \color{blue} 0 & \color{red} 1 \\
\color{red} 1 & \bf \color{green} 2 & \color{blue} 0 \\
\end{bmatrix}.
$$
However, the cyclic Latin square (generalizing the Latin square on the right above) has no orthogonal mate when $n$ is even (this goes back to Euler), so the desired Latin square impossible when $n$ is even.
Every broken diagonal and every broken antidiagonal contain every symbol
Possible if and only if $n \equiv \pm 1 \pmod 6$.
Since these are a special case of the previous case, they can only exist when $n$ is odd.
The diagonally cyclic Latin squares in DrLemniscate's answer work when $n \equiv \pm 1 \pmod 6$: along the main broken diagonals (and in particular, along the main diagonal) the symbols increase by $3$ (since $3$ generates $\mathbb{Z}_n$ in these cases).
When $n \equiv 3 \pmod 6$, it wasn't immediately obvious so I asked about it on MathOverflow. Afterwards, I realized it would give rise to a "strong complete mapping" and they don't exist when $n \equiv 3 \pmod 6$.
The main diagonal and the main antidiagonal contain every symbol
Possible if and only if $n \not\in \{2,3\}$.
Here, no conditions are applied to the non-main broken diagonals and antidiagonals. These are called doubly diagonal Latin squares. Julian Rosen's answer covers this, although there were other proofs around the same time:
A. J. W. Hilton, On double diagonal and cross latin squares,J. London Math, 1973 (link).
C. C. Lindner, On constructing doubly diagonalized latin squares, Feriodica Mathematica Hungarica Vol. 5 (3), (1974), pp. 249--253 (link).
You're unlikely to get a better write-up of the general case than the published papers on this. However, we've basically already solved a large number of cases.
If we have two doubly diagonal Latin squares of orders $n$ and $m$ and take their direct product (if $L_{ij}$ and $M_{ij}$ are the Latin squares, than its direct product is a new Latin square with $(L_{ij},M_{ij})$ in cell $(i,j)$), we get a doubly diagonal Latin square of order $nm$.
The diagonally cyclic Latin square construction when $n \equiv \pm 1 \pmod 6$ works here (which includes all primes $\geq 5$). And Hilton (op. cit.) gives these examples for orders $n \in \{4,6,8,9\}$.

So this is enough for when $n$ is not $2P$ or $3P$ where $P$ is a product of (not necessarily distinct) primes $\geq 5$. These cases seem to require more care, so it's best to refer to these papers.
The main diagonal contains every symbol
Possible if and only if $n \neq 2$.
I include this mostly for completeness sake. We take a diagonally cyclic Latin square of odd order $\geq 3$, and prolong it using a broken diagonal other than the main diagonal.
$$
\begin{bmatrix}
4 & \color{blue} 3 & 2 & 1 & 0 \\
1 & 0 & \color{blue} 4 & 3 & 2 \\
3 & 2 & 1 & \color{blue} 0 & 4 \\
0 & 4 & 3 & 2 & \color{blue} 1 \\
\color{blue} 2 & 1 & 0 & 4 & 3 \\
\end{bmatrix}
\longrightarrow
\begin{bmatrix}
4 & \color{blue} 5 & 2 & 1 & 0 & \color{red} 3 \\
1 & 0 & \color{blue} 5 & 3 & 2 & \color{red} 4 \\
3 & 2 & 1 & \color{blue} 5 & 4 & \color{red} 0 \\
1 & 4 & 3 & 2 & \color{blue} 5 & \color{red} 1 \\
\color{blue} 5 & 2 & 0 & 4 & 3 & \color{red} 2 \\
\color{red} 2 & \color{red} 3 & \color{red} 4 & \color{red} 0 & \color{red} 1 & \color{red} 5 \\
\end{bmatrix}
$$
When we prolong, we shift the contents of the chosen broken diagonal to both (a) a newly created column on the right, and (b) a newly created row at the bottom; fill the cells of the chosen broken diagonal with a new symbol; and add the new symbol to the bottom-right cell.