Does there exist a nonzero ring homomorphism from the ring of square rational matrices to the ring of rational numbers?

Notice that, as $\Bbb Z$-module, $M_n(\Bbb Q)$ is generated by the matrices of rank $1$, all of which must necessarily be mapped to $0$ (provided $n\ge2$). This is the case, for instance, because if $\operatorname{rk}A=1$, then there are invertible matrices $L$ and $R$ such that $LAR^{-1}$ is nilpotent. Therefore $0$ is the only multiplicative and additive map $M_n(\Bbb Q)\to \Bbb Q$. It might be worth mentioning that it is not a homomorphism of unital rings, because it does not map $1$ to $1$.

Yes, if $n=1$: the identity homomorphism.

Otherwise, no.

$M_n(\mathbb Q)$ is simple, so any nonzero ring homomorphism leaving it is injective.

But then $M_n(\mathbb Q)$ has many zero divisors if $n>1$, and those would have to map to zero divisors in $\mathbb Q$, of which there are $0$ or $1$, depending on how you like to count.

Here is yet another argument. Let $\theta:M _n (\mathbb Q)\to \mathbb Q $ be linear. It is straightforward to check then that $\theta=\operatorname {Tr}(A\cdot) $ for some $A\in M _n (\mathbb Q)$.

If $\theta $ is multiplicative, then in particular $\theta (BC)=\theta (B)\theta (C)=\theta (CB) $ for all $B,C $. Then $$ \operatorname {Tr}(ABC)=\operatorname {Tr}(ACB)=\operatorname {Tr}(BAC). $$ So $\operatorname {Tr}((AB-BA)C)=0$ for all $B,C $. Taking $C=(AB-BA)^T $ we obtain $AB-BA =0$. So $A $ commutes with all matrices, making it a scalar multiple of the identity. Thus $\theta $ is a scalar multiple of the trace; for $n\geq2$ it is easy to check that it can only be multiplicative if $A=0$.