Let me start things off by providing an upper bound. The bound is
very large, however, and I expect that it can be improved, perhaps
dramatically. But at least it shows the consistency of your large
cardinal relative to some other well-studied large cardinals.

**Theorem.** If $\kappa$ is $1$-$C^{(2)}$-extendible, then it is
forcing $\Pi^m_n$-indescribable for every $m,n$.

**Definition.** A cardinal $\kappa$ is
$1$-$C^{(2)}$-extendible,
if there is an elementary embedding $j:V_{\kappa+1}\to
V_{\theta+1}$, with critical point $\kappa$, such that the target
$j(\kappa)=\theta$ is $\Sigma_2$-correct in $V$, meaning
$V_\theta\prec_{\Sigma_2} V$.

This is a fairly strong large cardinal notion, far stronger than
the totally indescribable cardinals you mention in your question.
For example, every 1-extendible cardinal is superstrong and much
more. But these cardinals are weaker than Vopěnka's principle.

**Proof.** Assume that $\kappa$ is $1$-$C^{(2)}$-extendible. So
there is an elementary embedding $j:V_{\kappa+1}\to V_{\theta+1}$
with critical point $\kappa$ and $j(\kappa)=\theta$ is
$\Sigma_2$-correct in $V$.

Suppose now that $A\subset V_\kappa$ and $\langle
V_{\kappa+m},\in,A\rangle\models_f\varphi$, which means that
$\langle V[G]_{\kappa+m},\in,A\rangle\models\varphi$ for every
forcing extension $V[G]$. This is a $\Sigma_2$ property about $A$
and $\kappa$, since any violation of it would be revealed inside
some large enough $V_\eta$, using forcing inside that $V_\eta$.

Thus, by $\Sigma_2$-correctness, we see that $V_\theta$ agrees that
$\langle V_{\kappa+m},\in,A\rangle\models_f\varphi$. Since
$A=j(A)\cap\kappa$, we may pull this back by elementarity to
conclude that there is some $\lambda<\kappa$ with $\langle
V_{\lambda+m},\in,A\cap V_{\lambda+m}\rangle\models_f\varphi$
inside $V_\kappa$. But $\kappa$ itself must also be
$\Sigma_2$-correct, and so actually $\langle
V_{\lambda+m},\in,A\cap V_{\lambda+m}\rangle\models_f\varphi$ in
$V$, as desired. **QED**

I'll think some more about lower bounds and about pulling down the
strength of the hypothesis.

**Update.** I've realized that we can improve the upper bound as
follows. We don't really need the "$+1$", since that actually
provided a uniform version of the phenomenon, with the same
embedding working for every $A$.

Thomas Johnstone and I defined that a cardinal $\kappa$ is
*uplifting*, if it is inaccessible and $V_\kappa\prec V_\theta$ for
cofinally many inaccessible cardinals $\theta$. (J. D. Hamkins, T.
Johnstone, Resurrection axioms and uplifting
cardinals)
A boldface version is that $\kappa$ is *strongly uplifting*, if for
every $A\subset V_\kappa$ there are cofinally many inaccessible
cardinals $\theta$ for which $\langle
V_\kappa,\in,A\rangle\prec\langle V_\theta,\in,A^*\rangle$ for some
$A^*\subset V_\theta$. (J. D. Hamkins, T. Johnstone, Strongly
uplifting cardinals and boldface
resurrection)
These have diverse equivalent formulations, as I mention on the
linked blog post, connected with strengthenings of the strongly
unfoldable cardinals.

Let me now strengthen this a little more, for the present
application, with the following new large cardinal concept.

**Definition.** A cardinal $\kappa$ is *strongly
$C^{(n)}$-uplifting*, if for every $A\subset V_\kappa$ there is a
$\Sigma_n$-correct cardinal $\theta$ and $A^*\subset V_\theta$ with
$\langle V_\kappa,\in,A\rangle\prec\langle
V_\theta,\in,A^*\rangle$.

This is what we really needed in the theorem above.

**Theorem.** If $\kappa$ is strongly $C^{(2)}$-uplifting, then it
is forcing $\Pi^m_n$-indescribable for every $m$ and $n$.

**Proof.** Argue as in the first theorem above, but now we have
only $\langle V_\kappa,\in,A\rangle\prec\langle
V_\theta,\in,A^*\rangle$, instead of $j$. If $\langle
V_\kappa,\in,A\rangle+m\models_f\varphi$, then this will be true
inside $V_\theta$ since it is $\Sigma_2$-correct, and so $V_\theta$
thinks that this holds on an initial segment of $A^*$, and so we
get $\lambda<\kappa$ with $\langle V_\lambda,\in,A\cap
V_\lambda\rangle\models_f\varphi$ inside $V_\kappa$, which is right
about this since $\kappa$ is itself $\Sigma_2$-correct. **QED**

I think the strongly $C^{(2)}$-uplifting cardinals are
comparatively weak, and absolute to $L$, but I'll think more about
it.

isn'ttrue in all forcing extensions of $V$ ($A$ doesn't change but $V_\kappa$ does), so $V_\kappa\not\models_f$ it. $\endgroup$ – Noah Schweber Oct 14 '16 at 19:5510more comments