Height of diamond Assume $V=L$.
Let $\alpha$ be the least ordinal such that there is a $\Diamond_{\omega_1}$-sequence in $L_\alpha$.
It's obvious that $\omega_1 < \alpha < \omega_2$.
Do we have some better estimates of $\alpha$?
Also, what about $\Diamond^+$, squares and morasses?
 A: It depends on what diamond sequence you have.
$V=L$ proves there is a $\diamondsuit$-sequence of $L$-rank $\omega_1+1$: recall the famous way of proving the existence of $\diamondsuit$-sequence. The definition goes as follows (for example, Theorem 13.21 of Jech - Set theory (3rd edition)):

$(S_\alpha,C_\alpha)$ is the $<_L$-least pair $S_\alpha\subseteq\alpha$, $C_\alpha\subseteq\alpha$ a club, and $S_\alpha\cap\xi\neq S_\xi$ for all $\xi\in C_\alpha$. Otherwise $S_\alpha=C_\alpha=\alpha$.

Observe that if $X\subseteq\alpha<\omega_1$, then $(\alpha,X)\in L_{\omega_1}$. Furthermore, $<_L$ is absolute between $L$ and $L_{\omega_1}$. Combining with that $L_{\omega_1}$ satisfies $\mathsf{ZFC}^-$, which implies $L_{\omega_1}$ is closed under recursively defined sequences over there, we can see that the above definition works over $L_{\omega_1}$. Hence the $\diamondsuit$-sequence defined as above is a member of $\operatorname{Def}(L_{\omega_1})=L_{\omega_1+1}$.
However, the $L$-rank of a $\diamondsuit$-sequence can be arbitrarily large for a silly reason. Observe that the $(\alpha+1)$-th component of $\diamondsuit$-sequence does not affect being $\diamondsuit$, and hence we may replace it arbitrarily.
Let $f:\omega_1\to\omega_1$ be the increasing enumeration of successor ordinals below $\omega_1$. (Note that $f\in L_{\omega_1+1}$.) Recall the $\diamondsuit$-sequence $\langle A_\alpha\mid \alpha<\omega_1\rangle$ we previously defined (so of $L$-rank $\omega_1+1$.) For any $B\subseteq \omega_1$, define $A^B_\alpha$ by
$$A^B_{f(\alpha)}=\begin{cases} 0&\text{if }\alpha\in B \\ 1 & \text{if } \alpha\notin B\end{cases}$$
and $A^B_\alpha=A_\alpha$ for a limit $\alpha$. Then we can recover $B$ from $\langle A^B_\alpha\mid\alpha<\omega_1\rangle$ (possibly except for information about $0\in B$ and $1\in B$, but it does not matter.) Hence the $L$-rank of $\langle A^B_\alpha\mid\alpha<\omega_1\rangle$ is greater than or equal to that of $B$. (I suspect they are equal.)
There is no reason to believe that the $L$-rank of a $\diamondsuit$-sequence is equal to a given ordinal less than $\omega_2$. I think the question A tree coding all of $L_\alpha$ within $L_\alpha\cap\mathcal{P}(\omega)$ in MSE indicates, at least, the above construction does result in a $\diamondsuit$-sequence whose $L$-rank is equal to a given ordinal.
I believe the same argument applies for $\diamondsuit^+$-sequences, but I have not checked in detail. I am not sure the same holds for $\square$-sequences or morasses.
