No, second order logic does not have the weak test property: let $\mathfrak{B}=(\mathbb{R},{<})$ (that is, the real numbers with the only predicate being the usual "less than" order) and let $\mathfrak{A}=(\mathbb{R}\backslash\{0\},{<})$. Then $\mathfrak{B}\models$"I am a complete linear order" (completeness as in "for every $<$-downward closed set $X$ such that $X\neq\mathbb{R}$, there is a least upper bound for $X$, and likewise symmetrically"), whereas $\mathfrak{A}$ does not satisfy this, so $\mathfrak{A}\not\equiv_{\mathrm{SOL}}\mathfrak{B}$, and hence $\mathfrak{A}\not\preccurlyeq_{\mathrm{SOL}}\mathfrak{B}$. But property 1 does hold for $(\mathfrak{A},\mathfrak{B})$. For for simplicity let's first consider the case that the arity of $\varphi$ is 1. Let $x_1<x_2<\ldots<x_n$ be elements of $\mathbb{R}\backslash\{0\}$. Then the only subsets $X\subseteq\mathbb{R}$ which are second-order definable over $(\mathbb{R},{<})$ from $(x_1,\ldots,x_n)$ are finite unions of intervals with endpoints in $\{-\infty,x_1,\ldots,x_n,\infty\}$, and therefore, if $0\in X$, then there is an non-empty open interval $(-\varepsilon,\varepsilon)\subseteq X$, so $\varphi^{\mathfrak{B}}$ and $\varphi^{\mathfrak{B}}\cap\mathfrak{A}$ both have cardinality continuum. (E.g. if $\varepsilon>0$ is small enough then for each $x\in(-\varepsilon,\varepsilon)$ we can produce an automorphism $\pi:\mathfrak{B}\to\mathfrak{B}$ which fixes $x_1,\ldots,x_n$ but with $\pi(0)=x$.) For arity $k>0$ it is similar, with $k$-dimensional rectangles.
Yes, and a proper class of Woodin cardinals suffices. For $n<\omega$ and $X$ a set of ordinals, $M_n(X)$ denotes the minimal iterable proper class model $M$ of ZFC with $n$ Woodin cardinals above $\mathrm{rank}(X)$ and $X\in M$.
Because we have a proper class of Woodin cardinals, $M_n(X)$ exists for every set $X$ of ordinals. If $x$ is a real, then for $n$ even, $M_n(x)$ is boldface-$\Sigma^1_{n+2}$-correct,
and for $n$ odd, is boldface-$\Sigma^1_{n+1}$-correct. All forcing below is set forcing.
Claim 1: Given a structure $\mathfrak{B}$ and a set $B$ of ordinals coding $\mathfrak{B}$, and given an element $x\in\mathfrak{B}$ and a second order formula $\varphi$, if "$\mathfrak{B}\models\varphi(x)$" is forcing absolute then for all sufficiently large $n<\omega$, $M_n(B)\models$"It is forced by $\mathrm{Coll}(\omega,\mathrm{rank}(\mathfrak{B}))$ that $\mathfrak{B}\models\varphi(x)$".
(Here '"$\mathfrak{B}\models\varphi(x)$" is forcing absolute' just means that
the truth value of "$\mathfrak{B}\models\varphi(x)$" is independent of which generic extension of $V$ we are working in, but here $\mathfrak{B}$ and $x$ are fixed.) (Note I'm not saying that the converse of the claim holds.)
Proof: Let $G$ be $\mathrm{Coll}(\omega,\mathrm{rank}(\mathfrak{B}))$-generic over $V$, hence also over $M_n(B)$. Then $V[G]\models$"$\mathfrak{B}\models\varphi(x)$".
Since $\mathfrak{B}$ is countable in $V[G]$, this is just a projective statement there, about some reals in $M_n(B)[G]$. But one can show that (with our large cardinal assumption) $M_n(y)$ is not changed by forcing, so $(M_n(B))^V=(M_n(B))^{V[G]}$, and $V[G]\models$"$M_n(B)[G]=M_n(B,G)$", so $M_n(B)[G]$ is at least boldface-$\Sigma^1_{n+1}$-correct in $V[G]$. So for sufficiently large $n$,
$M_n(B)[G]\models$"$\mathfrak{B}\models\varphi(x)$", as desired.
Now working in $V$, let $\pi:H\to V_\gamma$ be elementary, with $\gamma$ a sufficiently large ordinal, and $H$ transitive and countable, and $\mathrm{rg}(\pi)$ containing $B,\mathfrak{B}$. The proper class mice $M_n(B)$ are determined by set-sized mice $M_n^\#(B)$, and these are definable from $B$ over $V_\gamma$,
so are also in $\mathrm{rg}(\pi)$. Let $\pi(\bar{B},\bar{M}_n^\#(\bar{B}))=(B,M_n^\#(B))$. Then because $\pi$ is elementary, in fact $\bar{M}_n^\#(\bar{B})=M_n^\#(\bar{B})$, i.e. the collapse of $M_n^\#(B)$ is the true $M_n^\#(\bar{B})$.
Claim 2: $\pi\upharpoonright\bar{\mathfrak{B}}:\bar{\mathfrak{B}}\to\mathfrak{B}$ is elementary with respect to forcing absolute second order formulas.
Proof: Suppose $\bar{\mathfrak{B}}\models\varphi(\bar{x})$ where $\varphi$ is forcing absolute, but $\mathfrak{B}\models\neg\varphi(x)$ where $x=\pi(\bar{x})$. Clearly $\neg\varphi$ is also forcing absolute, so in particular, "$\mathfrak{B}\models\neg\varphi(x)$" is forcing absolute, so by Claim 1, for sufficiently large $n$, $M_n(B)$ models 'it is forced by $\mathrm{Coll}(\omega,\mathrm{rank}(\mathfrak{B}))$ that $\mathfrak{B}\models\neg\varphi(x)$'. So $M_n(\bar{B})$ models the same about $\bar{\mathfrak{B}}$ and $\bar{x}$ (as the theory gets encoded into the $\#$ versions). We have an $(M_n(\bar{B}),\mathrm{Coll}(\omega,\mathrm{rank}(\bar{B})))$-generic $g\in V$ (as $M_n^\#(\bar{B})$ is countable). So for large $n<\omega$, $M_n(\bar{B})[g]$ models "$\bar{\mathfrak{B}}\models\neg\varphi(\bar{x})$". But $M_n(\bar{B})[g]=M_n(\bar{B},g)$, so this model is boldface-$\Sigma^1_{n+1}$ correct, so with $n$ large enough, this gives that $\bar{\mathfrak{B}}\models\neg\varphi(\bar{x})$, a contradiction.
Best Answer
In response to Noah's request after my earlier comments: this doesn't answer the question, but may shed some light. If $\kappa$ is supercompact, then for every $C$-satisfiable sentence $\phi$, there is a topological space $\mathcal{X}$ of size $<\kappa$ such that $\mathcal{X} \models_C \phi$. In fact, the entire powerset of (the underlying set of) $\mathcal{X}$---in particular, its topology---can be taken to be of size $<\kappa$.
Proof sketch: suppose $\mathcal{Y}=(Y,\tau) \models_C \phi$. Fix a regular $\theta$ such that $\mathcal{Y} \in H_\theta$, where $H_\theta$ denotes the collection of sets of hereditary cardinality $<\theta$ (you could work with rank initial segments if you prefer).
Since $\kappa$ is supercompact, there is some $\mathfrak{N} \prec (H_\theta,\in)$ of size $<\kappa$ such that $\mathcal{Y} \in \mathfrak{N}$ and the transitive collapse $\mathfrak{H}(\mathfrak{N})$ of $\mathfrak{N}$ is equal to $H_\mu$, where $\mu$ is the ordertype of $\mathfrak{N} \cap \theta$.
(This is folklore; see Fact 7.7 of my paper for a proof. In fact existence of such models capturing all possible sets is equivalent to supercompactness, by a classic theorem of Magidor).
In particular, this ensures that $\mathfrak{H}(\mathfrak{N})$ is correct about lots of things and closed under lots of operations.
For each $b \in \mathfrak{N}$, let $b_{\mathfrak{N}}$ denote the image of $b$ under the transitive collapsing map. By elementarity of the collapsing map and the fact that the statement ``$\mathcal{Y} \models_C \phi$" is downward absolute from the universe to the structure $(H_\theta,\in)$, it follows that
Since $\phi$ is just a formula in a countable language, the transitive collapsing map fixes $\phi$, i.e., $\phi = \phi_{\mathfrak{N}}$. So
Now $\mathfrak{H}(\mathfrak{N})$, being of the form $H_\mu$, is correct about the quoted statement (this is standard stuff; in fact, in this particular case the quoted statement is even $\Pi_1$, and sets of the form $H_\mu$ are always $\Pi_1$ elementary in the universe).
And the space $\Big( Y_{\mathfrak{N}}, \tau_{\mathfrak{N}} \Big)$ is of size $\le |\mathfrak{H}(\mathfrak{N})|= |\mathfrak{N}|<\kappa$.