It would be easier to count the number of words that have $2$ or fewer vowels, and then subtract this number from the total number of words (which you have already computed).
I assume that by "$3$" or more vowels" you mean $3$ or more occurrences of vowels, so in particular a word with 2 e's, 2 i's, and the rest consonants qualifies.
How many words are there with no vowels? Clearly
$$21^7$$
if, as per usual convention, we agree that there are $5$ vowels.
How many words with $1$ vowel? Where the vowel occurs can be chosen in $\binom{7}{1}$ ways. For each of these ways, the vowel can be chosen in $5$ ways. And once you have done that, the consonants can be filled in in $21^6$ ways, for a total of
$$\binom{7}{1}(5)(21^6)$$
Finally, how many with $2$ vowels? The location of the vowels can be chosen in $\binom{7}{2}$ ways. Once this has been done, the actual vowels can be put into these places in $5^2$ ways. And then you can fill in the consonants in $21^5$ ways, for a total of
$$\binom{7}{2}(5^2)(21^5)$$
Add up the $3$ numbers we have obtained, subtract from $26^7$.
Our argument was a little indirect. We could instead find, using the same sort of reasoning, the number of words with $3$ vowels, with $4$ vowels, with $5$, with $6$, with $7$, and add up. This is only a little more work than the indirect approach. But any saving of work is helpful! Also, the indirect approach that was described lets us concentrate on pretty simple situations, the most complicated of which is the $2$ vowel case.
Remark: The calculation we have done is closely connected to the Binomial Distribution, and if you have already covered this, it may be the point of the exercise. So if you know about the Binomial Distribution, imagine the letters to be chosen at random. Then the number of patterns with $3$ or more vowels is the probability of $3$ or more vowels, multiplied by $26^7$.
Your mistake is in the arithmetic. What you think comes out to 2341 really does come out to 240.
$4^5=1024$, $3^5=243$, $2^5=32$, $1024-(4)(243)+(6)(32)-4=1024-972+192-4=1216-976=240$
Best Answer
You forgot that you can have strings with two copies of one letter, two of another, and one of the third. There are $3$ ways to choose which of the letters is used only once, and $5$ ways to place it in the string. There are then $\binom42=6$ ways to choose which two of the remaining four places get the two copies of the first of the other two letters in alphabetical order, for a total of $3\cdot 5\cdot 6=90$ strings, exactly the number that you’re missing.
Added: To do it with generating functions, let $a_n$ be the number of words of length $n$ over a $3$-letter alphabet that use each letter at least once. The first step is to find a recurrence for $a_n$. For brevity call such words good. Each good word of length $n-1$ can be extended to a good word of length $n$ in $3$ different ways, since each of the three letters can be added at the end of the word. In addition, each $(n-1)$-letter word that contains exactly two of the three letters can be extended to a good word in one way, by adding the missing letter. How many of these $(n-1)$-letter words are there? There are $3$ ways to choose which of the three letters is missing, and each of the $n-1$ positions in the word can contain either of the two remaining letters. However, we have to exclude the two possibilities that contain only one of the two letters, so there are $3(2^{n-1}-2)$ such words. This leads us to the recurrence
$$a_n=3a_{n-1}+3(2^{n-1}-2)\tag{1}\;.$$
Clearly $a_0=a_1=a_2=0$ and $a_3=3!=6$. However, if we assume that $a_n=0$ for $n<0$, $(1)$ requires a small adjustment to get $a_0$ and $a_1$ right: as it stands, it yields the values $-9/2$ and $-3$. To correct this, we add a couple of Iverson brackets to get
$$a_n=3a_{n-1}+3(2^{n-1}-2)+\frac92[n=0]+3[n=1]\tag{2}\;.$$
Now multiply through by $x^n$ and sum over $n$ to get the generating function $f(x)$:
$$\begin{align*}f(x)=\sum_na_nx^n&=3\sum_n\left(a_{n-1}+2^{n-1}-2+\frac32[n=0]+[n=1]\right)x^n\\ &=3\left(\sum_na_{n-1}x^n+\sum_n2^{n-1}x^n-2\sum_nx^n+\frac32+x\right)\\ &=3\left(x\sum_na_{n-1}x^{n-1}+\frac12\sum_n(2x)^n-\frac2{1-x}+\frac32+x\right)\\ &=3\left(x\sum_na_nx_n+\frac{1/2}{1-2x}-\frac2{1-x}+\frac32+x\right)\\ &=3xf(x)+\frac32\left(\frac{7x-3}{(1-x)(1-2x)}+2x+3\right)\\ &=3xf(x)+\frac{6x^3}{(1-x)(1-2x)}\;, \end{align*}$$
so $$(1-3x)f(x)=\frac{6x^3}{(1-x)(1-2x)}\;,$$ and
$$\begin{align*} f(x)&=\frac{6x^3}{(1-x)(1-2x)(1-3x)}\\ &=x^3\left(\frac3{1-x}-\frac{24}{1-2x}+\frac{27}{1-3x}\right)\\ &=3x^3\left(\sum_nx^n-8\sum_n2^nx^n+9\sum_n3^nx^n\right)\\ &=\sum_n\left(3-3\cdot 2^{n+3}+3^{n+3}\right)x^{n+3}\\ &=\sum_n\left(3-3\cdot 2^n+3^n\right)x^n\;. \end{align*}$$
Equating coefficients, we finally have $a_n=3^n-3\cdot 2^n+3$.