Question: I'm trying to find an efficient way to compute maximum of a signal using its DFT. More formally:

$$\max\left\{ \mathcal F^{-1}\left(X\_k\right)\right\}, X\_k\text{ is the DFT of the signal and } \mathcal F^{-1} \text{ is the IDFT}$$

The original signal $x(n)$ is real, and it has some noise when sometimes there are wide peaks.

I'm looking for a solution that is quicker than the actual IFFT since the signal is very long (but we have its DFT).

Answer: 
There is no way to compute the (exact) maximum of the time domain signal directly from its DFT (without doing an IDFT first). Without any further knowledge, the best you can do is

$$|x[n]|=\left|\frac{1}{N}\sum\_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\right|\le\frac{1}{N}\sum\_{k=0}^{N-1}|X[k]| $$

which gives you an upper bound. However, this upper bound is usually not very tight, especially for large values of the DFT length $N$.
Generally:

There can be no shortcut here – every single element of the (inverse or forward, doesn't matter) DFT needs every element of the input. So, the FFT is already pretty much as fast as you can go.

However, you say:

> 
> The original signal $x(n)$ is real, and it has some noise when sometimes there are wide peaks.
> 
> 
> 

Aha! That means that your $X$ is (hermitian) symmetric. Which means you might be able to pick an FFT algorithm that benefits from that knowledge and can omit quite a few calculation – but I don't think that'll change the general $\mathcal O(N \log N)$ complexity of the IFFT, nor the $\mathcal O(N)$ of finding the maximum.