$\let\mr\mathrm$There are several results in the literature stating that a certain class $C$ satisfies $C\nsubseteq\mr{SIZE}(n^k)$ for any $k$, and usually it is straightforward to pad them to show that any barely superpolynomially expanded version of $C$ is not in $\mr{P/poly}$.
Let me say that $f\colon\mathbb N\to\mathbb N$ is a superpolynomial bound if it is time-constructible and $f(n)=n^{\omega(1)}$. For example, $n^{\log\log\log\log n}$ is a superpolynomial bound. In fact, an instructive exercise shows that if $g(n)$ is any unbounded monotone computable function, there is a superpolynomial bound $f$ such that $f(n)\le n^{g(n)}$.
First, direct diagonalization shows that $\Sigma_4^P\nsubseteq\mr{SIZE}(n^k)$ for any $k$. The same argument gives:
If $f$ is any superpolynomial bound, then $\Sigma_4\text-\mr{TIME}(f(n))\nsubseteq\mr{P/poly}$.
Proof sketch: For any $n$, let $C_n$ be the lexicographically first circuit of size $2f(n)$ that computes a Boolean function in $n$ variables not computable by a circuit of size $<f(n)$. Then, the language $L$ defined by $x\in L\iff C_{|x|}(x)=1$ works.
A well-known improvement states that $S_2\mr P\nsubseteq\mr{SIZE}(n^k)$ for any $k$. Likewise,
If $f$ is any superpolynomial bound, then $S_2\text-\mr{TIME}(f(n))\nsubseteq\mr{P/poly}$.
Proof sketch: If not, then in particular $\mr{NP}\subseteq S_2\mr P\subseteq\mr{P/poly}$, hence $\mr{PH}=S_2\mr P$. By a padding argument, $\Sigma_4\text-\mr{TIME}(f(n))\subseteq S_2\text-\mr{TIME}(f(n))\subseteq\mr{P/poly}$, quod non.
Oblivious classes do even better. Taking into account the objection raised by Apoorva Bhagwat, let $\mr{NLin=NTIME}(n)$. Then $\mr{NLin}\cup O_2\mr P\nsubseteq\mr{SIZE}(n^k)$ for any $k$, and the same argument yields:
If $f$ is any superpolynomial bound, then $\mr{NLin}\cup O_2\text-\mr{TIME}(f(n))\nsubseteq\mr{P/poly}$.
Proof sketch: If $\mr{NLin\subseteq P/poly}$, then by padding, $\mr{NP\subseteq P/poly}$, which implies $\mr{PH}=O_2\mr P$. Then we proceed as before.
In fact, we do have $O_2\mr P\nsubseteq\mr{SIZE}(n^k)$ for all $k$ due to a recent result of Li, who proved that there is a single-valued $\mr{FS_2P}$ algorithm solving the range-avoidance problem (aka dWPHP or sWPHP, as it is known in bounded arithmetic literature). Given an input $w\in\{0,1\}^n$, we can apply the range-avoidance algorithm to the function that takes a circuit $\{0,1\}^{(k+1)\log n}\to\{0,1\}$ of size $n^k$, and outputs its truth table; this gives us a Boolean function $h\colon\{0,1\}^{(k+1)\log n}\to\{0,1\}$ that requires circuit size $n^k$, and we apply it to the first $(k+1)\log n$ bits of $w$. This yields an $S_2\mr P$ language, which is oblivious (i.e., $O_2\mr P$) because we only use $n$ (rather than $w$) to find $h$.
(Note, by the way, that $O_2\mr P\subseteq\mr{P/poly}$.)
Now, pretty much the same argument applied to functions $\{0,1\}^{\log f(n)}\to\{0,1\}$ shows:
- If $f$ is any superpolynomial bound, then $O_2\text-\mr{TIME}(f(n))\nsubseteq\mr{P/poly}$.
These arguments are spelled out in Gajulapalli, Li & Volkovich.
There are also results involving MA. The often mentioned result that $\mr{MA\text-EXP\nsubseteq P/poly}$ is an overkill. Santhanam proved
$$\mr{promise\text-MA\cap promise\text-coMA\nsubseteq SIZE}(n^k)$$
for any $k$, and a similar argument gives:
If $f$ is any superpolynomial bound, then
$$\mr{promise\text-MA\text-TIME}(f(n))\cap\mr{promise\text-coMA\text-TIME}(f(n))\nsubseteq\mr{P/poly}.$$
Proof sketch: By Santhanam’s Lemma 11 (which is a sharpened version of the standard fact that $\mr{PSPACE=IP}$ with a PSPACE prover), there is a PSPACE-complete language $L$ and a randomized poly-time oracle TM $M$ such that on input $x$, $M$ only asks oracle queries of length $|x|$; if $x\in L$, then $M^L(x)$ accepts with probability $1$; and if $x\notin L$, then for any oracle $A$, $M^A(x)$ accepts with probability $\le1/2$.
For a suitable monotone polynomial $p$, let $A=(A_{\mr{YES}},A_{\mr{NO}})$ be the promise problem defined by
$$\begin{align}
(x,s)\in A_{\mr{YES}}&\iff\exists\text{circuit }C\,\bigl(p(|C|+|x|)\le f(|s|)\land\Pr[M^C(x)\text{ accepts}]=1\bigr),\\
(x,s)\in A_{\rlap{\mr{NO}}\phantom{YES}}&\iff\forall\text{circuit }C\,\bigl(p(|C|+|x|)\le f(|s|)\to\Pr[M^C(x)\text{ accepts}]\le1/2\bigr).
\end{align}$$
Let $h(x)$ be a polynomial reduction of $L$ to its complement, and let $B=(B_{\mr{YES}},B_{\mr{NO}})$ be the promise problem
$$\begin{align}
(x,s)\in B_{\mr{YES}}&\iff(x,s)\in A_{\mr{YES}}\land(h(x),s)\in A_{\mr{NO}},\\
(x,s)\in B_{\rlap{\mr{NO}}\phantom{YES}}&\iff(x,s)\in A_{\mr{NO}}\land(h(x),s)\in A_{\mr{YES}}.
\end{align}$$
If $p(n)$ is chosen suitably large,
$$B\in\mr{promise\text-MA\text-TIME}(f(n))\cap\mr{promise\text-coMA\text-TIME}(f(n)).$$
So, let us assume for contradiction that $B$ has polynomial-size circuits, say, $B\in\mr{SIZE}(n^k)$. Let $s(n)$ denote the size of the smallest circuit computing $L$ on inputs of length $n$, and put $t(n)=f^{-1}(p(s(n)))$; more precisely,
$$t(n)=\min\{m:p(s(n))\le f(m)\}.$$
Then $x\mapsto(x,1^{t(n)})$ is a reduction of $L$ to $B$, thus $L\in\mr{SIZE}(t(n)^k)$, which means
$$s(n)\le t(n)^k.$$
But since $f$ is superpolynomial, we have $t(n)=s(n)^{o(1)}$. This gives a contradiction for $n$ sufficiently large.
If we prefer a result with a non-promise version of MA, Miltersen, Vinodchandran, and Watanabe proved
$$\mr{MA\text-TIME}(f(n))\cap\mr{coMA\text-TIME}(f(n))\nsubseteq\mr{P/poly}$$
for a half-exponential function $f$. We can improve it in two ways: first, it holds for $\tfrac1k$-exponential bounds for any constant $k$, and second, it holds for oblivious classes. Here, a $\tfrac1k$-exponential function is, roughly speaking, a function $f$ such that $\underbrace{f\circ\dots\circ f}_k=\exp$. See the Miltersen–Vinodchandran–Watanabe paper and references therein for the precise definition; it involves a well-behaved family of well-behaved functions $e_\alpha(x)$, $\alpha\in\mathbb R_+$, such that $e_0(x)=x$, $e_1(x)=e^x-1$, and $e_{\alpha+\beta}=e_\alpha\circ e_\beta$. Also, if $f(n)\le e_\alpha(\mr{poly}(n))$ and $g(n)\le e_\beta(\mr{poly}(n))$, then $f(g(n))\le e_{\alpha+\beta}(\mr{poly}(n))$. Then we have:
$\mr{OMA\text-TIME}(e_\alpha)\cap\mr{coOMA\text-TIME}(e_\alpha)\nsubseteq\mr{P/poly}$ for any $\alpha>0$.
Proof sketch: Assume otherwise. Fix an integer $k$ such that $1/k<\alpha$. Let me abbreviate
$$\mr{OcOMT}(f)=\mr{OMA\text-TIME}(\mr{poly}(f(\mr{poly}(n)))\cap\mr{coOMA\text-TIME}(\mr{poly}(f(\mr{poly}(n))).$$
By padding, we have
$$\tag{1}\mr{OcOMT}(e_{\beta+1/k})\subseteq\mr{SIZE}(e_\beta(\mr{poly}(n)))$$
for any $\beta\ge0$. Moreover, using e.g. Santhanam’s Lemma 11 above, we have the implication
$$\tag{2}\mr{PSPACE\subseteq SIZE}(e_\beta(\mr{poly}(n)))\implies\mr{PSPACE\subseteq OcOMT}(e_\beta).$$
Since trivially $\mr{PSPACE\subseteq OcOMT}(e_1)$, a repeated application of (1) and (2) shows $\mr{PSPACE\subseteq SIZE}(e_{(k-1)/k}(\mr{poly}(n)))$, $\mr{PSPACE\subseteq OcOMT}(e_{(k-1)/k})$, $\mr{PSPACE\subseteq SIZE}(e_{(k-2)/k}(\mr{poly}(n)))$, $\mr{PSPACE\subseteq OcOMT}(e_{(k-2)/k})$, and so on. After $k$ steps, we reach
$$\mr{PSPACE\subseteq P/poly}\qquad\text{and}\qquad\mr{PSPACE=OMA\cap coOMA}.$$
Using padding once more, we get
$$\mr{DSPACE}(e_{1/k})\subseteq\mr{OcOMT}(e_{1/k})\subseteq\mr{P/poly},$$
which contradicts the results above, as $e_{1/k}$ is a superpolynomial bound.
