The answer is no: there is no such property. Any property of the form "a.s. something" that implies convergence in probability also implies a.s. convergence, hence cannot be equivalent to convergence in probability.
Proof:
Write $X=(X_n)_{n\in\mathbb{N}}$. Let's assume all variables $X_n$ are
binary (in $\{0;1\}$) for simplicity. Then convergence in probability
(implicitly to 0 in all that follows) of $X$ is simply:
$$\lim_{n\rightarrow+\infty} P(X_n=1)=0$$
For any property $\psi$ of sequences of 0s and 1s, "a.s. $\psi(X)$"
means "for almost all $\omega$, $\psi(X(\omega))$".
For every infinite
part $A$ of $\mathbb{N}$ define the property (of sequences of 0s and
1s):
$$\phi_A(x): \exists n\in A\quad x_n=0$$
Clearly, convergence in probability of $X$ implies a.s. $\phi_A(X)$.
Now assume convergence in probability of $X$ is implied by a.s.
$\phi(X)$ for some property $\phi$.
For any $A$, a.s. $\phi(X)$ implies convergence in probability of $X$
implies a.s. $\phi_A(X)$. Then it is clear that for any sequence $x$
of 0s and 1s, $\phi(x)$ implies $\phi_A(x)$: just use a random
sequence $X$ identically equal to $x$. Thus $\phi(x)$ implies $\forall A\space \phi_A(x)$ where $A$ ranges over all infinite parts of
$\mathbb{N}$. $\forall A\space \phi_A(x)$ says there is no infinite
part of $\mathbb{N}$ where $x$ is constant to 1. In other words it
says $x$ has only finitely many 1s. That is $x$ tends to 0. Thus a.s.
$\phi(X)$ implies a.s. convergence of $X$.
About about this property:
$$\phi((x_n)_{n\in\mathbb{N}}): \forall\epsilon>0\quad\lim_{n\rightarrow+\infty}\frac{\#\{k<n\mid |x_k|>\epsilon\}}{n}=0$$
"Almost surely $\phi$" is not equivalent to convergence in probability.
First, it is not implied by convergence in probability. A counter example is not so easy to find. The implication holds for independent variables. A counter example can be constructed by creating successive independent groups of 0/1 variables. All variables in the same group are equal: their common value is called the value of the group. And we have:
- the length of each group is large enough (say $2^k$) so that a group having value 1 influences the average sufficiently
- almost surely infinitely many groups have value 1 so that the above happens infinitely many times
- the probability for a group to have value 1 tends to 0 so that convergence in probability (to 0) holds
The last two points can be be obtained by saying that the probability for the $k^{th}$ group to have value 1 is $1/k$ and using the second Borrel-Cantelli lemma.
"Almost surely $\phi$" does not imply convergence in probability either. But it's not far from it. "For all subsequences of $X$, almost surely $\phi$ for this subsequence" implies convergence in probability and this is quite easy to prove.
Going further than this seems to raise key theoretical difficulties: how to define asymptotic empirical frequency in a more general way than this.
