First, ignoring all flushes (we do not need to distinguish between flushes and straight flushes), we will consider each of the other hand ranks that can be concurrent with a straight:
- No Pair
- One Pair
- Two Pair
- Three-of-a-Kind
With no pair, we can have a $7$-card straight ($8$ possible), a $6$-card straight with disjoint rank for the $7$th card ($47$ possible), or a $5$-card straight with distinct disjoint ranks for the $6$th and $7$th card ($162$ possible). In total we have $8+47+162=\color{blue}{217}$ possible $7$-card hands that contain a straight and no pair. For each of these, there are $4^7=16384$ ways to choose the suits. Counting the number of flushes, we find $1$ way to have $7$ cards in suit, $\binom76\cdot3=21$ ways to have $6$ cards in suit, and $\binom75\cdot3^2=189$ ways to have $5$ cards in suit, for a total of $211\cdot4=844$ flushes. This leaves $16384-844=\color{blue}{15540}$ non-flushes.
$$15540\times217=3372180\text{ straights with no pair}$$
With one pair, we can have a $6$-card straight (9 possible) or a $5$-card straight with a disjoint rank for the $6$th card (62 possible). For each, there are $6$ choices for the paired card, for a total of $71\cdot6=\color{blue}{426}$ possible $7$-card hands that contain a straight and one pair. For each of these, there are $\binom42\cdot4^5=6144$ ways to choose the suits. Counting the number of flushes, we find $3$ ways to have $6$ cards in suit and $3+\binom54\cdot3^2=48$ ways to have $5$ cards in suit, for a total of $51\cdot4=204$ flushes. This leaves $6144-204=\color{blue}{5940}$ non-flushes.
$$5940\times426=2530440\text{ straights with one pair}$$
With two pairs, we can only have a $5$-card straight (10 possible) with $2$ of the cards paired. There are a total of $10\cdot\binom52=\color{blue}{100}$ possible $7$-card hands that contain a straight and two pairs. For each of these, there are $\binom42^2\cdot4^3=2304$ ways to choose the suits. Counting the flushes, there are $3^2=9$ ways to choose the second suits for the paired cards, for a total of $36$ flushes. This leaves $2304-36=\color{blue}{2268}$ non-flushes.
$$2268\times100=226800\text{ straights with two pairs}$$
With three-of-a-kind, we can only have a $5$-card straight (10 possible) with one of the ranks repeated on the $6$th and $7$th card. There are a total of $10\cdot5=\color{blue}{50}$ possible $7$-card hands that contain a straight and two pairs. For each of these, there are $4^5=1024$ ways to choose the suits. Counting the flushes, for each suit there are $3$ ways to choose the missing suit for the set, for a total of $12$ flushes. This leaves $1024-12=\color{blue}{1012}$ non-flushes.
$$1012\times50=50600\text{ straights with three-of-a-kind}$$
Finally, adding these together give the total:
$$\begin{array}{r}3372180\\2530440\\226800\\+\quad50600\\\hline6180020\end{array}$$
Best Answer
First we count the number of hands with three of a kind, no four of a kind, and no other pair. There are $13$ ways to pick the three of a kind, $4$ ways to pick the three cards of that kind, $48$ acceptable cards for the first odd card, then $44,40,36$ for the following ones, but we have to divide by $4!=24$ for the orders of picking the four odd cards. That gives $$13\cdot 4 \cdot 48 \cdot 44 \cdot 40 \cdot 36/24=6,589,440$$ hands with three of a kind and no full house or four of a kind. Note this is not many more than the Wikipedia result. We now need to deduct the number of hands that have a straight or flush.
The flushes are reasonably easy. Again there are $13\cdot 4$ ways to get the three of a kind. Then there are $3$ ways to choose the suit of the flush and ${12 \choose 4}=495$ ways to select the other cards for a total of $77,220$ hands with three of a kind and a flush (which includes the straight flushes).
For straights we will start with the straight. The Wikipedia page shows (assuming that ace low straights count) there are $10,240$ ways to choose the five cards of a straight including straight flushes. Then there are $5$ ways to pick which card will have the three of a kind and $3$ ways to pick the two other cards, but we have to divide by $3$ for which of the three of a kind was part of the original straight so $10,240\cdot 5 \cdot 3/3=51,200$ hands to deduct.
We have deducted the straight flush hands twice, once for the straight and once for the flush. We need to add them in once. There are $40$ straight flushes, $5$ ways to pick the rank that has three of a kind, and $3$ ways to pick the missing suit for $600$
This winds up with $6,589,440-77,220-51,200+600=6,461,620$ hands with three of a kind and nothing higher, in agreement with the Wikipedia page.