It seems to me a sequence that is almost convergent implies that its Cesàro means converges but not vice versa.
What is the example that a not almost convergent sequence whose Cesàro means converge. i.e. for a sequence $\{x_i\}_{k \in \mathbb N}$
There exists an $L$, for all $\epsilon$, for all $n$, there exists a $p_0(n)$, such that for all $p > p_0(n)$, we have $$\left\lvert \frac{\sum_{k=1}^{p} x_{n+k-1}}{p}-L \right\rvert < \epsilon$$ and that $\lim_{n \to \infty}p_0(n) = +\infty$
On
Note that in the proper definition of almost convergence we can chose $n=0$ to get $$\forall \epsilon > 0 \exists N_0 \forall N > N_0 \quad \left|\frac1N \sum_{i=1}^N x_i - L\right| <\epsilon$$ wich is precisely the definition of a converging sequence for the sequence of Cesàro means $\sigma_N := \frac1N \sum_{i=1}^N x_i$
For the interesting question (the reversal of your original question), I think something like $x_i = (-1)^i \sqrt i$ should work. The key is to let the partial sums grow slowly, but the shifted partial sums (from almost convergence) unbounded due to the growth of $|x_i|$.
We get $$S_n = \sum_{i=1}^n (-1)^i \sqrt i \approx (-1, .41, -1.32, .68, \ldots)$$ And $\sigma_n \to 0$, but almost convergence should be violated.
Let $\{x_n\}_n$ be the $\pm1$ sequence $$\underbrace{+1}_1,\underbrace{-1,-1}_2,\underbrace{+1,+1,+1}_3,\underbrace{-1,-1,-1,-1}_4,\underbrace{+1,+1,+1,+1,+1}_5,{-1,\ldots} $$ i.e, the $n$th block of constant signs has length $n$ (a more explicit description of a somewhat similar sequence would be $x_n=(-1)^{\lfloor\sqrt n\rfloor}$). Then the sum of the first $n$ terms is $O(\sqrt n)$, hence the Cesáro limit is $0$. On the other hand, there are arbitrarily long consecutive blocks of $+1$ and of $-1$, respectively. Plugging such a block of length $\ge p$ into the definition of almost convegence $\to L$ gives both $|1-L|=\left|\frac{1+\ldots+1}p-L\right|<\epsilon$ and $|-1-L|=\left|\frac{(-1)+\ldots+(-1)}p-L\right|<\epsilon$, contradiction.