Can FFT information be used to obtain the DTFT?
Your question is a bit difficult to comprehend. It seems that you are requesting a closed form solution to the DTFT given the DFT of a sequence. This is of course possible. If there is no time-domain nor frequency-domain aliasing, then we can simply apply the formula for the IDFT on the samples (or equivalently perform the IFFT). Following this, we can apply the formula for DTFT on the time-domain samples. This may not be an answer in the simplified form you want, but it is a closed form expression for the DTFT and can be evaluated at any point. Furthermore, a very good computer algebra system (e.g., Maple, Mathematica, etc.) may be able to reduce it to the form that you have requested (meeting your requirement for no computation by hand). 

The DFT is a frequency-domain sampled version of the DTFT. The FFT computes the DFT. Because of this, you could simply apply the Shannon-Nyquist Sampling Theorem and interpolate the points of the DFT using sinc functions,

$$\DeclareMathOperator{\sinc}{sinc} 
\begin{align} X_{DTFT}(f) &amp;= \sum_{k=-\infty}^{\infty} X_{DFT}[k] \sinc(f - k) \\
&amp;= \sum_{k=0}^{N-1} X_{DFT}[k] \frac{\sin(\pi(f-k))}{N \sin\left(\tfrac{\pi}{N}(f-k)\right)}
\end{align}$$

where $\sinc(x) = \frac{\sin(\pi x)}{\pi x}$. This expression is equal to your simplified expression (i.e., it evaluates to the same value at every point). This expression assumes a unit length sampling interval (i.e., $T = 1$). This at least suggests a method for interpolating the DTFT from the DFT samples.

Looking for an exact simplified expression is difficult because different schools of thought will simplify it in different ways. Who is to say that your expression involving trigonometric functions is any better than one involving complex exponentials? For example, isn't the expression
$$ X_{DTFT}(f) = \frac{1}{4} + \frac{1}{4} \exp(-j 2 \pi f) + \frac{1}{4} \exp(-j 4 \pi f) + \frac{1}{4} \exp(-j 8 \pi f)$$
exactly what you are after? This is exactly what we would get if we used the approach in the first paragraph. Take the IFFT of the DFT, and then apply the formula for the DTFT. A computer program could do all of this and could evaluate this expression at any point $f$, requiring no computations by hand.