Is using 7-8 random words from all words of a language as password a good idea? The Bitcoin Bip-39 dictionary has 2048 words. And a random 24 of them are pretty secure together. What if the source is bigger (all words in the language) but there are only about 7-8 words? These words will be totally randomly selected by their index.
The Bitcoin Bip-39 dictionary with 2048 words can create $\approx 2^{263}$-entropy by tossing the coin 256 times to choose the words randomly. The random bits are converted into 24 11-bit blocks and every block is mapped into one of the 2048 words by ID. Since the random choice allows repetition, we need $n^r$ not $P(n,r)$;

$$2048^{24} \approx 2.9642774844752946\mathrm{e}{+79} \approx 2^{264}$$

If we assume that English has 171,476 words. Then with 8 words;

$$ 171476^{8}  \approx  7.475267765296064\mathrm{e}{+41} \approx 2^{139}.$$ Therefore you will have lower entropy the Bip-39, again. With 7 words;

$$ 171476^7   \approx  4.3593667716158903\mathrm{e}{+36} \approx 2^{121}.$$ Therefore you will have lower entropy than the Bip-39.  

As one can see, the choice 2048 and 24 in Bip-39 is clearly designed to reach $2^{264}$ entropy. If you use 16 words as passphrase ( as noted by Dan Neely) 

$$ 171476^{16}   \approx 5.587962816287441\mathrm{e}{+83} \approx 2^{278}.$$ that will exeed the Bip-39. 15 words reaches $3.258743390496303\mathrm{e}{+78} \approx 2^{260}$.


  Is it still considered secure


Yes, it is still secure and beyond the reach of any collective power, like the total power of bitcoin miners that can reach $2^{92}$ double SHA256 hashes in a year.



Note: I used to round a lot. I've decided to use full numbers since that changed at least one bit. So, here the Python;



import math

def prinPowInBase2(b,p):

    print(b,"^",p)
    print("\t", math.pow(b,p))
    print("\t", math.log2(math.pow(b,p)+1))
    print("\t", math.floor(math.log2(math.pow(b,p)+1)))

prinPowerInBase2(2048,24)