The answer is no.

Firstly, let's get the obvious out of the way. The Feferman–Levy model is a symmetric extension in which $\omega_1$ is singular. If we force the axiom of choice, we must collapse the ordinal to be countable.

Right. So perhaps we want to require that all successor cardinals are regular. This may be enough? But it is not going to be enough.

Gitik, Moti; Koepke, Peter, Violating the singular cardinals hypothesis without large cardinals, Isr. J. Math. 191, Part B, 901-922 (2012). ZBL1291.03094.

In that paper, the authors start with a model of $\sf GCH$, for any $\lambda>\aleph_{\omega+1}$, they build a symmetric extension which does not change cardinals and cofinalities of ordinals, where $\aleph^*(\mathcal P(\omega_\omega))>\lambda$. This is a violation of the Singular Cardinals Hypothesis which will require the existence of large cardinals, had it taken place in $\sf ZFC$.

Therefore, constructing the Gitik–Koepke model, where there is no inner model with the necessary large cardinal hypotheses, will result in a model where any extension by forcing to restore choice must collapse cardinals.

To your second question, Fernengel and Koepke had extended the original Gitik–Koepke model, first to a proper class situation, then for any set-many cardinals with $\sf DC_{<\lambda}$, under some additional constraints, which we can even show must violate Silver's theorem, and therefore collapse cardinals.

But, if $\sf DC_{<\lambda}$ holds, then at the very least we can force $\sf AC$ with a $\lambda$-closed forcing and therefore not add any bounded subsets to $\lambda$, thus preserving all cardinals up to $\lambda$ itself.

To see that, note that if $A$ is the seed for $\sf SVC$, and $\sf DC_{<\lambda}$ holds but $\sf DC_\lambda$ fails, then $\aleph(A)\geq\lambda$. Simply force with $A^{<\lambda}$, collapsing $A$ to have cardinality $\lambda$.

The answer is no.

Firstly, let's get the obvious out of the way. The Feferman–Levy model is a symmetric extension in which $\omega_1$ is singular. If we force the axiom of choice, we must collapse the ordinal to be countable.

Right. So perhaps we want to require that all successor cardinals are regular. This may be enough? But it is not going to be enough.

Gitik, Moti; Koepke, Peter, Violating the singular cardinals hypothesis without large cardinals, Isr. J. Math. 191, Part B, 901-922 (2012). ZBL1291.03094.

In that paper, the authors start with a model of $\sf GCH$, for any $\lambda>\aleph_{\omega+1}$, they build a symmetric extension which does not change cardinals and cofinalities of ordinals, where $\aleph^*(\mathcal P(\omega_\omega))>\lambda$. This is a violation of the Singular Cardinals Hypothesis which will require the existence of large cardinals, had it taken place in $\sf ZFC$.

Therefore, constructing the Gitik–Koepke model, where there is no inner model with the necessary large cardinal hypotheses, will result in a model where any extension by forcing to restore choice must collapse cardinals.

To your second question, Fernengel and Koepke had extended the original Gitik–Koepke model, first to a proper class situation, then for any set-many cardinals with $\sf DC_{<\lambda}$, under some additional constraints, which we can even show must violate Silver's theorem, and therefore collapse cardinals.

But, if $\sf DC_{<\lambda}$ holds, then at the very least we can force $\sf AC$ with a $\lambda$-closed forcing and therefore not add any bounded subsets to $\lambda$, thus preserving all cardinals up to $\lambda$ itself.

To see that, note that if $A$ is the seed for $\sf SVC$, and $\sf DC_{<\lambda}$ holds but $\sf DC_\lambda$ fails, then $\aleph(A)\geq\lambda$. Simply force with $A^{<\lambda}$, collapsing $A$ to have cardinality $\lambda$.