Most of the time, you don't actually need to calculate the state, but, sure, it's helpful when you're learning the topic to relate back to things you already know about.
The key ingredients that you have is that you state is a $+1$ eigenstate of a set of mutually commuting $\bar Z$ which all satisfy $\bar Z^2=I$. This tells you that the eigenvalues of $\bar Z$ are $\pm 1$. We can therefore construct a projector onto the +1 eigenstate by simplify evaluating $(I+\bar Z)/2$. The projector onto our state is then just $$ |\psi\rangle\langle\psi|=\prod_i(I+\bar Z_i)/2. $$$$ |\psi\rangle\langle\psi|=\prod_i\left((I+\bar Z_i)/2\right). $$ So, one option that you have is just to take $$ \frac14(I-YY)(I+ZX) $$ and find its eigenvector with +1 eigenvalue.
Alternatively, just guess a state $|\phi\rangle$. If you calculate $$ (|\psi\rangle\langle\psi|)|\phi\rangle=\langle\psi|\phi\rangle\ |\psi\rangle, $$ then so long as you got lucky enough that $\langle\psi|\phi\rangle\neq 0$, the output you get is proportional to the state you're trying to identify. For example, I could take $|\phi\rangle=|00\rangle$: \begin{align*} (I-YY)(I+ZX)|00\rangle&=(I-YY)(|00\rangle+|01\rangle) \\ &=|00\rangle+|01\rangle-(-|11\rangle+|10\rangle) \\ &=|00\rangle+|01\rangle-|10\rangle+|11\rangle. \end{align*}