On p.112-113 of volume 10-1 of Gauss's werke, which contain an unpublished fragment dated to 1805, Gauss states some results on cubic and biquadratic "Gaussian periods" in a trigonometric form. Here is a Google translation of this fragment:
Cubic: Let $4p = a^2+27b^2$, so that: $\frac{a}{\sqrt{4p}} = cos\varphi, \frac{b\sqrt{27}}{\sqrt{4p}} = sin\varphi$. Then the period of order $\frac{1}{3}(p-1)$ is: $$-\frac{1}{3}+\frac{2}{3}cos\frac{1}{3}\varphi \sqrt{p}$$ Biquadratic: Let $p = a^2+4b^2$, so that: $\frac{a}{\sqrt{p}} = cos\varphi, \frac{2b}{\sqrt{p}} = sin\varphi$. Then the period of order $\frac{1}{4}(p-1)$ is: $$-\frac{1}{4}+\frac{1}{4}\sqrt{p}+\frac{1}{2}\sqrt{p}\cdot cos\frac{1}{2}\varphi = \prod$$, and: $${\prod}^0 + i\prod'-\prod''- i\prod''' = \sqrt{p}\cdot (\frac{a+b\sqrt{-4}}{a-b\sqrt{-4}})^{\frac{1}{4}}$$
Just to make things clearer: $p$ is a prime number, and Bachmann (the commentor on Gauss's number theoretical works) remarked that the result for the cubic case is connected to article 358 of Gauss's Disquisitions Arithmeticae.
I believe that, except the last result, the trigonometric form of the cubic and biquadratic "Gaussian periods" can be explained by someone who is familiar with the theory of Gauss sums. However, the last result is unclear since Gauss doesn't define ${\prod}^0, \prod',\prod'',\prod'''$. Therefore, the first priority in considering this question is explaining the meaning of those symbols. In addition, any useful comment about the trigonometric form of the Gaussian periods will be blessed!
I checked in article 358 of the D.A and I saw that the result for the cubic case already appears in Gauss's footnote to this article, but with different notation:
Note that here Gauss denotes the prime number by $4n = M^2+27N^2$ (the prime number is $n$), and the Gaussian period by $p$.
In light of this footnote to article 358 of D.A, i believe i can give an answer to the notational aspects of my own question. I think i understand what is the meaning of ${\prod}^0, \prod',\prod'',\prod'''$, for the biquadratic case. There are $p-1$ invertible residues modulo $p$, of which exactly $(p-1)/4$ are quartic residues, and $3(p-1)/4$ are quartic non-residues. The quartic residues form a subgroup of index 4 of the group of invertible residues. Therefore, i think that ${\prod}^0$ is the Gauss sum over the quartic residues, $\prod'$ is the Gauss sum over the first coset of the subgroup of quartic residues, $\prod''$ its second coset and $\prod'''$ its third coset. Those four Gauss sums are weighted by the four units in the Gaussian integers $+1,+i,-1,-i$, and then summed up. Obviously i have no idea how to get the result
$${\prod}^0 + i\prod'-\prod''- i\prod''' = \sqrt{p}\cdot (\frac{a+b\sqrt{-4}}{a-b\sqrt{-4}})^{\frac{1}{4}}$$
Note also that this result is analogous to the one that appears in the footnote, for the cubic case; $p+\epsilon p'+{\epsilon}^2p'' = (n(M+N\sqrt{-27})/2)^{\frac{1}{3}}$.
My answer clarified what is meant in Gauss's result, but i still didn't find sources that discuss the particular result of Gauss for the biquadratic case (Gauss's footnote to art. 358, which treats the cubic case, is already discussed in Lemermeyer's book on reciprocity laws). Therefore, i'll be glad if someone will give a source that discusses this result.