You are quite right that it is possible to consider efficiency also for estimators which are not asymptotically normal.
Consider for example estimating the upper bound of the uniform distribution $U[0,\theta]$. It is well known that the sample maximum $y_{(n)}$ is the MLE, for a random sample $y$.
$y_{(n)}$ is not unbiased so let us bias-correct it. Its cdf is
\begin{eqnarray*}
F_{y_{(n)}}(x)&=&\Pr\{Y_1\leqslant x,\ldots,Y_n\leqslant x\}\\
&=&\Pr\{Y_1\leqslant x\}^n\\
&=&\begin{cases}
0&\qquad\text{for}\quad x<0\\
\left(\frac{x}{\theta}\right)^n&\qquad\text{for}\quad 0\leqslant x\leqslant\theta\\
1&\qquad\text{for}\quad x>\theta
\end{cases}
\end{eqnarray*}
Thus, its density is
$$f_{y_{(n)}}(x)=
\begin{cases}
\frac{n}{\theta}\left(\frac{x}{\theta}\right)^{n-1}&\qquad\text{for}\quad 0\leqslant x\leqslant\theta\\
0&\qquad\text{else}
\end{cases}
$$
Hence,
\begin{eqnarray*}
E[Y_{(n)}]&=&\int_0^\theta x\frac{n}{\theta}\left(\frac{x}{\theta}\right)^{n-1}dx\\
&=&\int_0^\theta n\left(\frac{x}{\theta}\right)^{n}dx\\
&=&\frac{n}{n+1}\theta
\end{eqnarray*}
Thus, an unbiased estimator is given by $$\frac{n+1}{n}y_{(n)},$$
as
$$E\left[\frac{n+1}{n}Y_{(n)}\right]=\frac{n+1}{n}\frac{n}{n+1}\theta=\theta$$
Its variance is
\begin{eqnarray*}
Var\left(\frac{n+1}{n}Y_{(n)}\right)&=&\left(\frac{n+1}{n}\right)^2Var(y_{(n)})\\
&=&\left(\frac{n+1}{n}\right)^2\left[E(Y^2)-\left(\frac{n}{n+1}\theta\right)^2\right]\\
&=&\left(\frac{n+1}{n}\right)^2\left[\int_0^\theta n\left(\frac{x^{n+1}}{\theta^{n}}\right)dx-\left(\frac{n}{n+1}\theta\right)^2\right]\\
&=&\left(\frac{n+1}{n}\right)^2\left[\frac{n}{n+2}\frac{\theta^{n+2}}{\theta^n}-\left(\frac{n}{n+1}\theta\right)^2\right]\\
&=&\left(\frac{n+1}{n}\right)^2\left[\frac{n}{n+2}\theta^2-\left(\frac{n}{n+1}\theta\right)^2\right]\\
&=&\left(\frac{n+1}{n}\right)^2\left[\frac{n(n+1)^2\theta^2-n^2(n+2)\theta^2}{(n+2)(n+1)^2}\right]\\
&=&\frac{1}{n(n+2)}\theta^2
\end{eqnarray*}
Another unbiased estimator is given by twice the sample mean. It has variance
$$Var(2\bar X)=\theta^2/n$$
So we see that the variance of the MLE converges to zero at a faster rate, $O(n^{-2})$ than that of twice the sample mean, $O(n^{-1})$, so that, so to speak , the MLE is infinitely more efficient asympotically.
This may be motivated by the fact that the asymptotic distribution of the MLE is not normal, see e.g. here.
