Pfaffian representation of the Fermat quintic
Edit Jan 11, 2018: One can make a little bit of progress by pushing my initial idea (pair partitions plus round robin) "to second order", i.e., make use of a little bit of cancellation. See further below.
Edit Jan 10, 2018:
There is a flaw in the reasoning below as pointed out by Will.
One needs more conditions on the combinatorial design like the only way to produce a pair partition of $[10]$ only using pairs in $\cup_i P_i$ is to take one the $P_i$'s. I don't know if this extra requirement can be satisfied. Perhaps a round robin tournament scheduling algorithm might work.
Initial version (faulty!):
It follows from the existence of five set partitions $P_1,\ldots,P_5$ of the set $[10]:=\{1,2,\ldots,10\}$ into pairs such that these partitions are disjoint. Namely a pair $\{i,j\}\subset [10]$ with $i\neq j$ cannot be used by more than one partition.
Let us assume this existence. Then for each $P_i$ define its "standard permutation" as $\sigma_i$ given by listing the pairs by increasing order of its minimum and writing each pair in increasing order. Let $\epsilon_i$ be the sign of this standard permutation. We now define a tensor $T=(T_{i,j,k})_{(i,j,k)\in [10]\times[10]\times[5]}$ as follows. If a pair $\{i,j\}\in P_k$ with $i<j$, set $T_{i,j,k}=\epsilon_k$ and $T_{j,i,k}=-\epsilon_k$. Otherwise make all the other tensor entries zero. Define the matrix of linear forms $M_{i,j}(x)=\sum_{k=1}^{5}T_{i,j,k} x_k$. A moment of thought will show you that the Pfaffian of this matrix is a multiple of the Fermat quintic.
Fiddling around produces for instance this choice of partitions:
$$ P_1 =\{\ \{1,2\},\ \{3,4\},\ \{5,6\},\ \{7,8\},\ \{9,10\}\ \} $$ $$ P_2 =\{\ \{1,10\},\ \{2,3\},\ \{4,5\},\ \{6,7\},\ \{8,9\}\ \} $$ $$ P_3 =\{\ \{1,6\},\ \{2,7\},\ \{3,8\},\ \{4,9\},\ \{5,10\}\ \} $$ $$ P_4 =\{\ \{1,3\},\ \{2,10\},\ \{4,6\},\ \{5,8\},\ \{7,9\}\ \} $$ $$ P_5 =\{\ \{1,5\},\ \{2,6\},\ \{3,9\},\ \{4,7\},\ \{8,10\}\ \} $$
Edit Jan 11, 2018 (continued):
1) The Fermat quadratic $F=x_1^2+x_2^2$.
My construction works as is and one can take for the partitions $P_1$ and $P_2$ any two of the three pair partitions of $[4]$. To do this in a more principled way, let me modify the standard round robin scheduling algorithm in the even case. Instead of fixing $1$ and turning everybody else, I will turn $1$ too. This gives the two arrays $$ P_1=\left( \begin{array}{cc} 1 & 2 \\ 4 & 3 \end{array} \right)\ \ ,\ \ P_2=\left( \begin{array}{cc} 4 & 1 \\ 3 & 2 \end{array} \right)\ . $$ The pairs correspond to the columns of the arrays. The corresponding signs are $\epsilon_1=\epsilon_2=1$. This gives the matrix $$ M=\left( \begin{array}{cccc} 0 & x_2 & 0 & x_1 \\ -x_2 & 0 & x_1 & 0 \\ 0 & -x_1 & 0 & x_2 \\ -x_1 & 0 & -x_2 & 0 \end{array} \right) $$ with ${\rm Pf}(M)=F$.
2) The Fermat cubic $F=x_1^3+x_2^3+x_3^3$.
The modified round robin rule gives the three pair partitions $$ P_1=\left( \begin{array}{ccc} 1 & 2 & 3 \\ 6 & 5 & 4 \end{array} \right)\ ,\ P_2=\left( \begin{array}{ccc} 6 & 1 & 2 \\ 5 & 4 & 3 \end{array} \right)\ ,\ P_3=\left( \begin{array}{ccc} 5 & 6 & 1 \\ 4 & 3 & 2 \end{array} \right)\ . $$ One can also use an array of pairs to see what happens: $$ \left( \begin{array}{ccc} (16) & (25) & (34) \\ (23) & (14) & (56) \\ (45) & (36) & (12) \end{array} \right) $$ The partitions $P_1$, $P_2$, $P_3$ correspond to the rows of this array and their signs are $\epsilon_1=\epsilon_2=\epsilon_3=1$.
A rather tedious inspection shows that the only pair partitions one can form using any pairs in this $3\times 3$ array are the previous ones and the three new ones corresponding to the columns of the array, i.e., $P_4=(16)(23)(45)$, $P_5=(25)(14)(36)$ and $P_6=(34)(56)(12)$. Remark that this looks nicer if one draws chord diagrams on a circle with three colors for $P_1$, $P_2$, $P_3$. The corresponding signs are $\epsilon_4=1$, $\epsilon_5=-1$ and $\epsilon_6=1$. Without any modification the Pfaffian should be $$ x_1^3+x_2^3+x_3^3+x_1 x_2 x_3-x_1 x_2 x_3+x_1 x_2 x_3 $$ where I emphasized the individual contributions of $P_1,\ldots,P_6$ on purpose. Now modify the weights of the pairs in the first row by factors $a,b,c$. The Pfaffian now becomes $$ abcx_1^3+x_2^3+x_3^3+ax_1 x_2 x_3-bx_1 x_2 x_3+cx_1 x_2 x_3 $$ so we get the Fermat cubic if we pick these weights so that $$ abc=1\ \ ,\ \ a-b+c=0 $$ For example we can take $a=\theta$, $b=1$, $c=\theta^5$ where $\theta=e^{\frac{i\pi}{3}}$.
More explicitly, the Fermat cubic is $F={\rm Pf}(M)$ with $$ M=\left( \begin{array}{cccccc} 0 & x_3 & 0 & x_2 & 0 & \theta x_1 \\ -x_3 & 0 & x_2 & 0 & x_1 & 0 \\ 0 & -x_2 & 0 & \theta^5 x_1 & 0 & x_3 \\ -x_2 & 0 & -\theta^5 x_1 & 0 & x_3 & 0 \\ 0 & -x_1 & 0 & -x_3 & 0 & x_2 \\ -\theta x_1 & 0 & -x_3 & 0 & -x_2 & 0 \end{array} \right)\ . $$
There is an obvious pattern here but I don't know if it works for $n=4$ or the OP's question, that is, $n=5$. In fact, I just realized (via performing the same permutation on rows and columns) these are determinantal cases as in the comment by Will.